The Cone Definition Theorems solved Exercised Problems Common Vertex Notations Intersection Of A Line With A Cone Reciprocal Cone

The Cone Enveloping Cone

Definition Of A Cone With Common Vertex And Examples

Definition. Let S be a surface and P be a point not on the surface. The set of tangent lines to the surface S and passing through P form a cone with vertex at P. This is called the enveloping cone or the tangent cone of the given surface.

Theorem.1. The enveloping cone of the sphere x² + y² + z² = a² with vertex at (x₁, y₁, z₁) is (xx₁ + yy₁ + zz₁ – a²)² = (x² + y² + z² – a²)(x₁² + y₁² + z₁² – a²).

Proof. Let S = x² + y² + z² – a² = 0

P(x₁, y₁, z₁) ∉ S = 0

⇒ x₁² + y₁² + z₁² – a² ≠ 0

Let Q(x, y, z) be a point on the enveloping cone C.

∴ \(\overleftrightarrow{\mathrm{PQ}}\) is a tangent line to the sphere S = 0.

Let R ∈ \(\overleftrightarrow{\mathrm{PQ}}\) and (R, P, Q) = λ : 1

∴ \(\mathrm{R}=\left[\frac{\lambda x+x_1}{\lambda+1}, \frac{\lambda y+y_1}{\lambda+1}, \frac{\lambda z+z_1}{\lambda+1}\right]\)

R ∈ S = 0

⇒ \(\left(\frac{\lambda x+x_1}{\lambda+1}\right)^2+\left(\frac{\lambda y+y_1}{\lambda+1}\right)^2+\left(\frac{\lambda z+z_1}{\lambda+1}\right)^2=a^2\)

⇒ \(\left(\lambda x+x_1\right)^2+\left(\lambda y+y_1\right)^2+\left(\lambda z+z_1\right)^2=a^2(\lambda+1)^2\)

⇒ \(\lambda^2\left(x^2+y^2+z^2-a^2\right)+2 \lambda\left(x x_1+y y_1+z z_1-a^2\right)+\left(x_1^2+y_1^2+z_1^2-a^2\right)=0\)

⇒ \(\lambda^2 S+2 \lambda S_1+S_{11}=0\)

If \(\overleftrightarrow{\mathrm{PQ}}\) is a tangent line to the sphere then the two roots of the equation (1) are equal

⇒ 4S₁² – 4SS₁₁ = 0 ⇒ s₁² = SS₁₁

Hence the equation to the enveloping cone C is s₁² = SS₁₁

i.e., (xx₁ + yy₁ + zz₁ – a²)² = (x² + y² + z² – a²)(x₁² + y₁² + z₁² – a²)

The Cone Right Circular Cone

Definition. A right circular cone is a surface generator by a line which passes through a fixed point, and makes a constant angle with a fixed line through the fixed point.

Let S be a set of concurrent lines, concurrent at V. If there exists a line L passing through V such that for a line M, M ∈ S ⇒ (L, M) = θ the S is called a right circular cone with vertex at V.

The line L is called the axis is θ the semi – vertical angle of the cone.

Note. The section of a right circular cone by any plane perpendicular to the axis is a circle.

Theorems On Cones With Solved Problems Step-By-Step

Theorem.2. The equation of a right circular cone with vertex at (α, β, γ), semi-vertical angle θ and axis having direction ratios (l, m, n) is [l(x – α) + m(y – β) + n(z – γ)]² = (l² + m² + n²)[(x – α)² + (y – β)² + (z – γ)²]cos²θ

Proof. Let V be the vertex and VL be the axis of the cone. V = (α, β, γ) and the direction ratios of the axis VL are (l, m, n).

Let P(x, y, z) be a point on the cone.

D.r’s of V.P are (x – α, y – β, z – γ)

Semi vertical angle θ = \((\overleftrightarrow{\mathrm{VL}}, \overleftrightarrow{\mathrm{VP}})\)

⇒ \(\cos \theta=\frac{l(x-\alpha)+m(y-\beta)+n(z-\gamma)}{\sqrt{\left[(x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2\right]} \sqrt{\left(l^2+m^2+n^2\right)}}\)

Hence the equation of the right circular cone is

⇒ \(\left[(x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2\right]\left(l^2+m^2+n^2\right) \cos ^2 \theta\)

= \([l(x-\alpha)+m(y-\beta)+n(z-\gamma)]^2\)

Corollary 1. If the vertex be the origin then the equation of the cone becomes (lx + my + nz)² = (l² + m² + n²)(x² + y² + z²)cos²θ

Corollary 2. The equation of the right circular cone with vertex at (0, 0, 0) and whose axis is the z-axis and semi-vertical angle α is x² + y²  = z² tan²α

Proof. Since d.r’s of the z-axis are (0, 0, 1)

l = 0, m = 0, n = 1

∴ The equation to the right circular cone is (x² + y² + z² ) cos²α = z²

⇒ (x² + y²) = z² (sec²α – 1) ⇒ (x² + y²) = z² (tan²α)

The Cone Solved Problems

Example.1. Find the enveloping cone of the sphere x² + y² + z² + 2x – 2y = 2, with its vertex at (1, 1, 1).

Solution.

Given Vertex = (1, 1, 1).

Equation to the given sphere is S = x² + y² + z² + 2x – 2y – 2 = 0

Now s₁ = x.1 + y.1 + z.1 + (x + 1) – (y + 1) – 2 = 0 = 2x + z – 2

S₁₁ = 1 + 1 + 1 + 2 – 2 – 2 = 1

∴ The equation to the enveloping cone is s₁² = SS₁₁

(2x + z – 2)² = (x² + y² + z² + 2x – 2y – 2)(1) ⇒ 3x² – y² + 4zx – 10x + 2y – 4z + 6 = 0

Intersection Of A Line With A Cone Examples And Solutions

Example.2. Find the equation to the right circular cone whose vertex is P(2, -3, 5), axis PQ which makes equal angles with the axis and which passes through A(1, -2, 3).

Solution.

Given

vertex is P(2, -3, 5)

A(1, -2, 3)

The axis of the cone makes equal angles θ with the coordinate axes

∴ d.r’s of the axis are (cosθ, cosθ, cosθ) ⇒ d.r’s of the axis are (1, 1, 1)

Let α be the semi-vertical angle of the cone with vertex P(2, -3, 5)

∴ The equation to the required cone is

[(x – 2)² + (y + 3)² + (z – 5)²] (1 + 1 + 1)cos²α = [1.(x – 2) + 1.(y + 3) + 1(z – 5)]²

The point A(1, -2, 3) lies on the cone

<=> \(\left[(1-2)^2+(-2+3)^2+(3-5)^2\right] 3 \cos ^2 \alpha=[(1-2)+(-2+3)+(3-5)]^2\) <=> [/latex]\cos \alpha=\frac{\sqrt{2}}{3}[/latex]

∴ The equation to the required cone is

⇒ \(\left[(x-2)^2+(y+3)^2+(z-5)^2\right] \times \frac{2}{3}=[(x-2)+(y+3)+(z-5)]^2\)

Simplifying the equation x² + y² + z² + 6(yz + zx + xy) – 16x – 36y – 4z – 28 = 0

Example.3. Find the equation of the right circular cone with vertex at (2, 1, -3) and whose axis is parallel to OY and whose semi-vertical angle is 45°.

Solution. Axis is parallel OY ⇒ d.cs. of axis are (0, 1, 0)

Given semi vertical angle α = 45°, vertex = (2, 1, -3).

∴ Equation to the cone is [(x – 2)² + (y – 1)² + (z + 3)²] =(y – 1)² + (z + 3)²](0 + 1 + 0)cos²45

= [0.(x – 2) + 1.(y – 1) + 0.(z + 3)]²

⇒ \(\frac{1}{2}\left[(x-2)^2+(y-1)^2+(z+3)^2\right]=(y-1)^2 \Rightarrow(x-2)^2-(y-1)^2+(z+3)^2=0\)

⇒ x² – y² + z² – 4x + 2y + 6z + 12 = 0

Example.4. Find the equation of the right circular cone whose vertex is the origin, axis as the line x = t, y = 2t, z = 3t and whose semi-vertical angle is 60°.

Solution. Vertex (α, β, γ) = (0, 0, 0)

Equation to the axis \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=t\)

⇒ D.r’s of the axis (l, m, n) = (1, 2, 2)

Semi vertical angle = 60°

∴ Equation to the required cone is

[(x – 0)² + (y – 0)² + (z – 0)²][1² + 2² + 3²]cos²60° = [1.(x – 0) + 2.(y – 0) + 3.(z – 0)]²

⇒ \(\frac{14}{4}\left(x^2+y^2+z^2\right)=(x+2 y+3 z)^2\)

⇒ 7(x² + y² + z²) = 2(x² + 4y² + 9z² + 4xy + 12yz + 6zx)

⇒ 5x² – y² – 11z² – 24yz – 12zx – 8xy = 0

Step-By-Step Solutions For Line And Cone Intersection Problems

Example.5. Show that the plane z = 0 cuts the enveloping cone of the sphere x² + y² + z² = 11 which has its vertex at (2, 4, 1) in a rectangular hyperbola.

Solution. Let S ≡ x² + y² + z² – 11 = 0

Given point P = (2, 4, 1) = (x₁, y₁, z₁)

S₁ ≡ xx₁ + yy₁ + zz₁ – 11 ≡ x(2) + y(4) + z(1) – 11 = 2x + 4y + z – 11

S₁₁ ≡ x₁² + y₁² + z₁² – 11 ≡ (2)² + (4)² + (1)² – 11 = 10

∴ Equation to the enveloping cone is SS11 = S12

⇒ (x² + y² + z² – 11)(10) = (2x + 4y + z – 11)²

Where the plane z = 0 cuts the cone, then the equation to the conic is

10(x² + y² – 11) = (2x + 4y – 11)² ⇒ 6x² – 6y² – 16xy + 88y + 44x – 331

In this equation coefficient of x² + coefficient of y² = 6 – 6 = 0

⇒ The conic is a rectangular hyperbola

Example.6. Find the equation of the cone generated by rotating the line \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) about the line \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) as axis.

Solution. Given lines pass through the origin ⇒ vertex is the origin.

D.r’s of axis are (a, b, c)

Semi-vertical angle = angle between the generator and the axis

⇒ \(\cos \theta=\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\) …..(1)

∴ Equation to the cone is [(x – 0)² + (y – 0)² + (z – 0)²](a² + b² + c²).cos²θ

= [a(x – 0) + b(y – 0) + c(z – 0)]²

Using (1) we have ⇒ (x² + y² + z²)(al + bm + cn)² = (l² + m² + n²)(ax + by + cz)²

Example.7. If α is the semi-vertical angle of a right circular cone which passes through the lines OY, OZ, and x = y = z. Show that cosα = (9 – 4√3)^-1/2.

Solution. Let (l, m, n) be d.r’s of the axis of the cone

D.r.’s of OY are (0, 1, 0)

D.r.’s of OZ are (0, 0, 1)

α is the angle between the axis and OY

⇒ \(\cos \alpha=\frac{0 . l+1 \cdot m+0 . n}{\sqrt{0+1+0} \sqrt{l^2+m^2+n^2}}=\frac{m}{\sqrt{l^2+m^2+n^2}}\) …..(1)

Also α is the angle between the axis and OZ

⇒ \(\cos \alpha=\frac{0 . l+0 . m+1 . n}{\sqrt{0+0+1} \sqrt{l^2+m^2+n^2}}=\frac{n}{\sqrt{l^2+m^2+n^2}}\) …..(2)

From (1) and (2) m = n

Similarly angle between the axis and \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)is

⇒ \(\cos \alpha=\frac{1 . l+1 \cdot m+1 \cdot n}{\sqrt{1+1+1} \sqrt{l^2+m^2+n^2}}=\frac{l+m+n}{\sqrt{3\left(l^2+m^2+n^2\right)}}\) …..(3)

Equating (1) and (3) \(m=\frac{l+m+n}{\sqrt{3}}\)

⇒ \(l+m(1-\sqrt{3})+n=0 \Rightarrow l+m(1-\sqrt{3})+m=0\) (∵ m = u)

⇒ \(l=m(\sqrt{3}-2) \Rightarrow \frac{l}{\sqrt{3}-2}=\frac{m}{1}=\frac{n}{1}\)

∴ From (1) \(\cos \alpha=\frac{1}{\sqrt{(\sqrt{3}-2)^2+1+1}}=\frac{1}{\sqrt{9-4 \sqrt{3}}}=(9-4 \sqrt{3})^{-1 / 2}\)

Solved Problems On Reciprocal Cones Step-By-Step

Example.8. Lines are drawn through the origin with direction ratios (1, 2, 2), (2, 3, 6), and (3, 4, 12). Find the direction ratios of the axis of the right circular cone and hence show that its semi-vertical angle is cos^-1(1/√3). Also, find the equation of the cone.

Solution.

Given

Lines are drawn through the origin with direction ratios (1, 2, 2), (2, 3, 6), and (3, 4, 12).

Let (l, m, n) be the direction ratios of the axis of the right circular cone

Let α be the semi-vertical angle of the cone

∴ Each given line is at α with the axis

(1) \(\cos \alpha=\frac{1 . l+2 \cdot m+2 \cdot n}{\sqrt{1+4+4} \sqrt{l^2+m^2+n^2}}\) …..(1)

(2) \(\cos \alpha=\frac{2 \cdot l+3 \cdot m+6 \cdot n}{\sqrt{4+9+36} \sqrt{l^2+m^2+n^2}}\) …..(2)

(3) \(\cos \alpha=\frac{3 \cdot l+4 \cdot m+12 \cdot n}{\sqrt{9+16+144} \sqrt{l^2+m^2+n^2}}\) …..(3)

From (1) and (2) : \(\frac{l+2 m+2 n}{3}=\frac{2 l+3 m+6 n}{7} \Rightarrow l+5 m-4 n=0\) …..1

From (1) and (3): \(\frac{1}{3}(l+2 m+2 n)=\frac{1}{13}(3 l+4 m+12 n) \Rightarrow 2 l+7 m-5 n=0\) …..2

Solving 1 and 2: \(\frac{l}{-25+28}=\frac{m}{-8+5}=\frac{n}{7-10} \Rightarrow \frac{l}{3}=\frac{m}{-3}=\frac{n}{-3} \Rightarrow \frac{l}{1}=\frac{m}{-1}=\frac{n}{-1}\)

∴ Direction cosines of the axis are \(\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)\)

∴ The semi-vertical angle is given by

From (1) \(\cos \alpha=\frac{1(1)+2(-1)+2(-1)}{\sqrt{1+4+4} \sqrt{1+1+1}}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Equation the cone is \(\cos \alpha=\frac{1}{\sqrt{3}}=\frac{1(x-0)-1(y-0)-1(z-0)}{\sqrt{1+1+1} \sqrt{x^2+y^2+z^2}}\)

⇒ (x – y – z)² = x² + y² + z² ⇒ yz – zx – xy = 0

Example.9. find the equation of the cone formed by rotating the line 2x + 3y = 6, z = 0 about the y – axis.

Solution.

Given line 2x + 3y = 6, z = 0

The direction cosines of the axis are (0, 1, 0)

Given equation to the generator is 2x + 3y = 6, z = 0

2x = -3(y – 2), z = 0

⇒ \(\frac{x}{3}=\frac{y-2}{-2}=\frac{z}{0}\) …..(1)

Also Y-axis meets the line 2x + 3y = 6, z = 0 at (0, 2, 0)

⇒ vertex of the plane = (0, 2, 0)

∴ Semi-vertical angle = Angle between the line (1) and Y – axis.

⇒ \(\cos \alpha=\frac{0.3+1(-2)+0.0}{\sqrt{0+1+0} \sqrt{9+4+0}}=\frac{-2}{\sqrt{13}}\)

∴ The equation to the right circular cone with vertex (0, 2, 0) and axis d.r.’s (0, 1, 0) is

[0(x – 0) + 1(y – 2) + 0.z]² (0 + 1 + 0)cos²α = (0.x + 1(y – 2) + 0.z)²

⇒ \([x+(y-2)+z]^2 \frac{4}{13}=(y-2)^2 \Rightarrow 4 x^2-9(y-2)^2+4 z^2=0\)

The Cone Notation

Let S represent the second-degree general equation in x, y, z. The following notation is used in this chapter.

i.e. S ≡ ax² + by² + cz²+ 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d

E = E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy

U = ax + hy + gz + u; V = hx + by + fz + v;

W = gx + fy + cz + w; D = ux + vy + wz + d and

U₁ = ax₁ + hy₁ + gz₁ + u; V₁ = hx₁ + by₁ + fz₁ + v;

W₁ = gx₁ + fy₁ + cz₁ + w; D₁ = ux₁ + vy₁ + wz₁ + d

Then S₁ = axx₁ + byy₁ + czz₁ + f(yz₁ + y₁z) + g(zx₁ + z₁x) + h(xy₁ + x₁y) + u(x + x₁) + v(y + y₁) + w(z + z₁) + d

= (ax₁ + hy₁ + gz₁ + u)x + (hx₁ + by₁ + fz₁ + v)y + (gx₁ + fy₁ + cz₁ + w) + ux₁ + vy₁ + wz₁ + d = U₁x + V₁y + W₁z + D₁

S₁₁ = U₁x₁ + V₁y₁ + W₁z₁ + D₁

Worked Examples Of Line-Cone Intersections With Solutions

Theorem.3. If (x₁, y₁, z₁) is the vertex of the cone S = 0 then U₁ = V₁ = W₁ = D₁ = 0

Proof. Let the equation to the cone be

S = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0

Given vertex of the cone, P = (x₁, y₁, z₁)

Shifting the origin to the point P the new equation of the cone referred to vertex P as the new origin is

a(x + x₁)² + b(y + y₁)² + c(z + z₁)² + 2f(y + y₁)(z + z₁) + 2g(x + x₁)(z + z₁) + 2h(x + x₁)(y + y₁) + 2u(x + x₁) + 2v(y + y₁) + d = 0

⇒ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2x(ax₁ + hy₁ + gz₁ + u) + 2y(hx₁ + by₁ + fz₁ + v) + 2z(gx₁ + fy₁ + cz₁ + w) + ax₁² + by₁² + cz₁² + 2fy₁z₁ + 2gz₁x₁ + 2hx₁y₁ + 2ux₁ + 2vy₁ + 2wz₁ + d = 0.

⇒ E(x, y, z) +2U₁x + 2V₁y + 2W₁z + S₁₁ = 0

This must be a homogeneous equation

⇒ U₁ = 0, V₁ = 0, W₁ = 0 and S₁₁ = 0.

But S₁₁ = U₁x₁ + V₁y₁ + W₁z₁ + D₁ = 0 ⇒ D₁ = 0 ⇒ U₁ = V₁ = W₁ = D₁ = 0

Corollary 1. If the equation S = 0 represents a cone then the condition is

⇒ \(\left|\begin{array}{llll}
a & h & g & u \\
h & b & f & v \\
g & f & c & w \\
u & v & w & d
\end{array}\right|=0\)

Proof. Eliminating x₁, y₁, z₁ in the equations

U₁ = ax₁ + hy₁ + gz₁ + u = 0; V₁ = hx₁ + by₁ + fz₁ + v = 0;

W₁ =gx₁ + fy₁ + cz₁ + w= 0

D₁ = ux₁ + vy₁ + wz₁ + d = 0

We get \(\left|\begin{array}{llll}
a & h & g & u \\
h & b & f & v \\
g & f & c & w \\
u & v & w & d
\end{array}\right|=0\)

This is the required condition that the equation S = 0 represents a cone.

Corollary 2. The vertex ( x₁, y₁, z₁ ) satisfies the equations

U ≡ ax + hy + bz + u = 0 …..(1) V ≡ hx + by + fz + v = 0 …..(2)

W ≡ gx + fy + cz + w = 0 …..(3) D ≡ ux + vy + wz + d = 0 …..(4)

Thus the vertex is obtained by solving any three of the above four equations

Note. Consider the homogeneous polynomial

S(x, y, z, t) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2uxt + 2vyt + 2wzt + dt2

Now \(\frac{\partial \mathrm{S}}{\partial x}=2(a x+h y+g z+u t)\)

⇒ \(\frac{\partial \mathrm{S}}{\partial y}=2(h x+b y+f z+v t) ; \quad \frac{\partial \mathrm{S}}{\partial z}=2(g x+f y+c z+w t) ; \quad \frac{\partial \mathrm{S}}{\partial t}=2(u x+v y+w z+d t)\)

Now equating \(\frac{\partial S}{\partial x}, \frac{\partial S}{\partial y}, \frac{\partial S}{\partial z}, \frac{\partial S}{\partial t}\) each to zero and putting t = 1, we get

U = V = W = D = 0.

The Cone Solved Problems

Example.1. If ax² + by² + cz² + 2ux + 2vy + 2wz + d = 0 represents a cone prove that \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=d\).

Solution.

Given

If ax² + by² + cz² + 2ux + 2vy + 2wz + d = 0 represents a cone

Let ( x₁, y₁, z₁ ) be the vertex of the given cone.

The given equation represents a cone if

⇒ \(\mathrm{U}_1=0 \Rightarrow a x_1+u=0 \quad \Rightarrow x_1=\frac{-u}{a}\);

⇒ \(\mathrm{V}_{\mathrm{I}}=0 \quad \Rightarrow b y_1+v=0 \Rightarrow y_1=\frac{-v}{b}\)

⇒ \(\mathrm{W}_1=0 \Rightarrow c z_1+w=0 \Rightarrow z_1=\frac{-w}{c}\) and

D₁ = 0 ⇒ ux₁ + vy₁ + wz₁ + d = 0

Substituting in D₁ = 0 the values of x_{1 +}  y_{1 +}  z₁ we get

⇒ \(u\left(\frac{-u}{a}\right)+v\left(\frac{-v}{b}\right)+w\left(\frac{-w}{c}\right)+d=0\)

⇒ \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=d\)

Properties Of Reciprocal Cones With Solved Exercises

Example.2. Find the vertex of the cone 7x² + 2y² + 2z² – 10zx + 10xy + 26x – 2y + 2z – 17 = 0

Solution.

Given

7x² + 2y² + 2z² – 10zx + 10xy + 26x – 2y + 2z – 17 = 0

Consider the homogeneous equation

S(x, y, z,t) = 7x² + 2y² + 2z² – 10zx + 10xy + 26xt – 2yt + 2zt – 17t2 = 0

∴ \(\frac{\partial \mathrm{S}}{\partial x}=14 x-10 z+10 y+26 t=14 x+10 y-10 z+26\) (∵ t = 1)

⇒ \(\frac{\partial S}{\partial y}=4 y+10 x-2 t=10 x+4 y-2\)

⇒ \(\frac{\partial \mathrm{S}}{\partial z}=4 z-10 x+2 t=-10 x+4 z+2\);

⇒ \(\frac{\partial \mathrm{S}}{\partial t}=26 x-2 y+2 z-34 t=26 x-2 y+2 z-34\)

Coordinates of vertex satisfy the equations

14x + 10y – 10z + 26 = 0 …..(1) 10x + 4y – 2 = 0 …..(2)

-10x + 4z + 2 =0 …..(3) 26x – 2y + 2z – 34 = 0 …..(4)

Solving (1), (2), and (3) we get x = 1, y = -2, z = 2

Substituting (1, -2, 2) in (4) 26 + 4 + 4 + – 34 = 0

Hence the vertex of the cone is (1, -2, 2)

Example.3. Show that the equation 2y² – 8yz – 4zx – 8xy + 6x – 4y – 2z + 5 = 0 represents a cone whose vertex is \(\left(-\frac{7}{6}, \frac{1}{3}, \frac{5}{6}\right)\)

Solution.

Given

2y² – 8yz – 4zx – 8xy + 6x – 4y – 2z + 5 = 0

Making the given equation homogeneous, we get

S(x, y, z, t) = 2y² – 8yz – 4zx – 8xy + 6xt – 4yt – 2zt + 5t2 = 0

⇒ \(\frac{\partial \mathrm{S}}{\partial x}=-4 z-8 y+6 t ; \quad \frac{\partial \mathrm{S}}{\partial y}=4 y-8 z-8 x-4 t\)

⇒ \(\frac{\partial \mathrm{S}}{\partial z}=-8 y-4 x-2 t ; \quad \frac{\partial \mathrm{S}}{\partial t}=6 x-4 y-2 z+10 t\)

Equating t = 1 coordinates of the vertex satisfies the equations

4y + 2z – 3 = 0 …..(1) 2x – y + 2z + 1 = 0 …..(2)

2x + 4y + 1 = 0 …..(3) 3x – 2y – z + 5 = 0 …..(4)

Solving (1),(2) and (3) we get \(x=-\frac{7}{6}, y=\frac{1}{3}, z=\frac{5}{6}\)

Substituting in (4): \(3\left(-\frac{7}{6}\right)-2\left(\frac{1}{3}\right)-\frac{5}{6}+5=0 \Rightarrow-21-4-5+30=0\)

Hence the vertex of the cone is \(\left(-\frac{7}{6}, \frac{1}{3}, \frac{5}{6}\right)\)

Theorem.4. The cone E(x, y, z) = 0 will have three mutually perpendicular generators <=> a + b + c = 0.

Proof. Given the equation of the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

Let \(\frac{x}{p}=\frac{y}{q}=\frac{z}{r}\) …..(1) be a generator of the cone,

∴ E(p, q, r) = 0 ⇒ ap² + bq² + cr² + 2fqr + 2grp + 2hpq = 0 …..(2)

The equation to the plane ⊥er to (1) and passing through the vertex is px + qy + rz = 0 …..(3)

Let this plane intersect the cone along two real generators and (l, m, n) be the d.c’s of one of the generators.

∴ al² + bm² + cn² + 2fmn + 2gnl + 2hlm = 0 …..(4)

pl + qm + rn = 0 …..(5)

Eliminating n between (4) and (5), we get

⇒ \(l^2\left(a r^2+c p^2-2 g r p\right)+2 l m\left(c p q+h r^2-g q r-f r p\right)+m^2\left(b r^2+c q^2-2 f q r\right)=0\)

⇒ \(\frac{l^2}{m^2}\left(a r^2+c p^2-2 g r p\right)+2 \frac{l}{m}\left(c p q+h r^2-g q r-f r p\right)+\left(b r^2+c q^2-2 f q r\right)=0\) …..(6)

If (l₁, m₁, n₁) and (l₂, m₂, n₂) are the direction cosines of the two generators of intersection then \(\frac{l_1}{m_1}, \frac{l_2}{m_2}\) are the roots of (6).

∴ \(\frac{l_1 l_2}{m_1 m_2}=\frac{b r^2+c q^2-2 f q r}{a r^2+c p^2-2 g r p}\)

⇒ \(\frac{l_1 l_2}{b r^2+c q^2-2 f g r}=\frac{m_1 m_2}{a r^2+c p^2-2 g r p}=\frac{n_1 n_2}{a q^2+b p^2-2 h p q}=k\), by symmetry.

∴ l₁l₂ + m₁m₂ + n₁n₂ = k[a(q² + r²) + b(r² + p²) + c(p² + q²) – 2fqr – 2grp – 2hpq]

= k(a + b + c)(p² + q² + r²) …..(2)

The two generators of the intersection of the plane (3) with the cone are at right angles.

<=> l₁l₂ + m₁m₂ + n₁n₂ = 0

<=> a + b + c = 0 [(∵ (p, q, r) ≠ (0, 0, 0)

Since plane (3) is perpendicular to generator (1), the two generators of intersection of the plane (3) with the cone are perpendicular to generator (1).

∴ These three generators are mutually perpendicular

<=> Two generators of intersection are perpendicular <=> a + b + c = 0.

Note. 1. Above condition is satisfied whatever is the direction of the generator. From this, we get if three mutually perpendicular lines are generators to the cone, then a + b + c = 0

Note. 2. Let F(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 be a cone.

Shifting the origin to the vertex the transformed equation is

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0.

the cone F(x, y, z) = 0, has three mutually perpendicular generators

<=> E(x, y, z) = 0 has three mutually perpendicular generators <=> a + b + c = 0

The Cone Solved Problems

Example.1. Show that the two lines of intersection of the plane ax + by + cz = 0 with the cone yz + zx + xy = 0 will be perpendicular if \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\).

Solution. Given cone is yz + zx + xy = 0.

In this equation Coefficient of x² + Coefficient of y² + Coefficient of z² = 0

∴ The cone contains sets of three mutually perpendicular generators.

The plane ax + by + cz = 0 cuts the cone in perpendicular generators if it’s a normal line through the vertex (0, 0, 0).

i.e., \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) is a generator of the cone.

⇒ bc + ca + ab = 0 ⇒ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Classification Of Cones By Vertex And Reciprocal Relations With Examples

Example.2. If the line \(x=\frac{1}{2} y=z\) represents one of the three mutually perpendicular generators of the cone 11yz + 6zx – 14xy = 0, find the equations of the other two.

Solution. The given cone is 11yz + 6zx – 14xy = 0

The plane through the vertex of the cone and perpendicular to the generator.

⇒ \(\frac{x}{1}=\frac{y}{2}=\frac{z}{1}\) …..(1) is x + 2y + z = 0 the other two generators perpendicular to (1) are the lines of intersection of 11yz + 6zx – 14xy = 0 and x + 2y + z = 0.

Let l, m, n be the direction ratios of one of the common lines.

Then 11mn + 6nl – 14lm = 0 …..(2)

and l + 2m + n = 0 ⇒ n = -1-2m

Substituting in (2) 11m(-l-2m) + 6l(-l -2m)-14lm = 0

⇒ 6l² + 37lm + 22m² = 0 ⇒ (2l + 11m)(3l + 2m) = 0

⇒ 2l + 11m = 0 or 3l + 2m = 0

(1) solving l + 2m + n = 0

2l + 11m + 0.n = 0

we get \(\frac{l}{-11}=\frac{m}{2}=\frac{n}{7}\)

(2) solving 1 + 2m + n = 0

3l + 2m + 0.n = 0

we get \(\frac{1}{-2}=\frac{m}{3}=\frac{n}{-4}\)

∴ the other two perpendicular generators are \(\frac{x}{-11}=\frac{y}{2}=\frac{z}{7} \text { and } \frac{x}{2}=\frac{y}{-3}=\frac{z}{4}\)

Example.3. Show that if a right circular cone has sets of three mutually perpendicular generators, its semi-vertical angle must be tan^-1√2.

Solution. Let the origin be the vertex, l, m, n be direction cosines of the axis of the cone and α be its semi-vertical angle

Then the equation to cone is (lx + my + nz)² = (l² + m² + n²)(x² + y² + z²)cos²α

∵ the cone contains three mutually perpendicular generators, then

i.e., Coefficient of x² + Coefficient of y² + Coefficient of z² = 0.

Coefficient of x² = l²-(l² + m² + n²)cos²α

Coefficient of y² = m²-(l² + m² + n²)cos²α

Coefficient of z² = n²-(l² + m² + n²)cos²α

Adding, we have by (1) (l² + m² + n²)-3(l² + m² + n²)cos²α = 0

⇒ 1 – 3cos²α = 0 ⇒ tan²α = 2 ⇒ tanα = √2 ⇒ α = tan^-1 √2

Example.4. Show that cone whose vertex is the origin and which passes through the curve of intersection of the surface 2x² – y² + 2z² = 3d² any plane a distance d, from the origin has three mutually perpendicular generators.

Solution. The equation to any plane at a distance d from the origin is lx + my + nz = d …..(1)

where l, m, n are the actual d.c’s of normal to the plane.

Homogenizing the equation of the sphere with that of the plane, we have

⇒ \(2 x^2-y^2+2 z^2=3 d^2\left(\frac{l x+m y+n z}{d}\right)^2\)

Now Coefficient of x² + Coefficient of y² + Coefficient of z²

= (2 – 3l²) – l – 3m² + (2 – 3n²) = 3 – 3(l² + m² + n²) = 3 – 3(1) = 0

Hence plane (1) cuts the cone in three mutually perpendicular generators.

Example.5. If the plane 2x – y + cz = 0 cuts the cone yz + zx + xy = 0 in perpendicular lines find c.

Solution. Given cone yz + zx + xy = 0 …..(1)

contains sets of three mutually perpendicular generators.

2x – y + cz = 0 cuts (1) in perpendicular lines

⇒ the normal of the plane lies on it.

⇒ (2, -1, c) must satisfy the cone equation

⇒ (-1)(c) + c(2) + (2)(-1) = 0 ⇒ c = 2

Example.6. Find the locus of the point from which three mutually perpendicular lines can be drawn to intersection the central conic ax² + by² = 1, z = 0.

Solution. Let the point P be (x₁, y₁, z₁)

Any line through P is \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=k\) …..(1)

Any point on the line is (x1 + lk, y1 + mk, z1 + nk)

the point lies on the base curve ax² + by² = 1, z = 0

<=> a(x₁ + lk)² + b(y₁ + mk)² = 1, z₁ + nk = 0

Eliminating k, we have \(a\left(x_1-\frac{l z_1}{n}\right)^2+b\left(y_1-\frac{m z_1}{n}\right)^2=1\)

⇒ a(nx₁ – lz₁)² + b(ny₁ – mz₁) = n²

using (1), to the cone is

a[x₁(z – z₁) – z₁(x – x₁)]² + b[y₁(z – z₁) – z₁(y – y₁)]² = (z – z₁)²

This contains three mutually perpendicular generators

<=> Coefficient of x² + coefficient of y²+ Coefficient of z² = 0 ⇒ az₁² + bz₁² + ax₁² + by₁² – 1 = 0.

∴ Locus of P is a(x² + z²) + b(y² + z²) = 1

The Cone Intersection Of A Line With A Cone

Let the equation to the cone S be

S(x, y, z) ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vu + 2wz + d = 0

Let the equation to a line be \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}(=r)\)

Let P be a point on this line

∴ p = (lr + x₁, mr + y₁, nr + z₁) = 0

∴ P ∈ s <=> S(lr + x₁, mr + y₁, nr + z₁) = 0

<=> \(a\left(l r+x_1\right)^2+b\left(m r+y_1\right)^2+c\left(n r+z_1\right)^2+2 f\left(m r+y_1\right)\left(n r+z_1\right)+2 g\left(n r+z_1\right)\left(l r+x_1\right)\)

⇒ \(+2 h\left(l r+x_1\right)\left(m r+y_1\right)+2 u\left(l r+x_1\right)+2 v\left(m r+y_1\right)+2 w\left(n r+z_1\right)+d=0\).

<=> \(r^2\left(a l^2+b m^2+c n^2+2 f m n+2 g m l+2 h l m\right)+2 r\left[l\left(a x_1+h y_1+g z_1+u\right)\right.\)

⇒ \(\left.+m\left(h x_1+b y_1+f z_1+v\right)+n\left(g x_1+f y_1+c z_1+w\right)\right]+\mathrm{S}\left(x_1, y_1, z_1\right)=0\)

<=> \(r^2 \mathrm{E}(l, m, n)+2 r\left[l \mathrm{U}_1+m \mathrm{~V}_1+n \mathrm{~W}_1\right]+\mathrm{S}_{11}=0\)

(1) This will be a quadratic equation in r <=> E(l, m, n) ≠ 0

The equation will have two real and dictinctive roots.

<=> (lU₁ + mV₁ + nW₁)² – E(l, m, n) S₁₁ > 0

Then there will be two real points of the line common with the cone. The line segment joining the two point is called the chord of the cone.

(2) If E(l, m, n) ≠ 0 and (lU₁ + mV₁ + nW₁)² – E(l, m, n) S₁₁ > 0 then there are no common points.

(3) If E(l, m, n) ≠ 0 and (lU₁ + mV₁ + nW₁)²   =  E(l, m, n) S₁₁ then the two roots of the equation are real and equal. Hence the line meets the curve in two coincident points. Then the line is called the tangent line ar that common point.

(4) If E(l, m, n) = lU₁ + mV₁ + nW₁ = S₁₁ = 0, then the line becomes a generator of the cone.

The Cone Tangent Plane

Definition. Let S = 0, be the cone and L be a tangent line to the cone at P on it. The locus of the line L is called the tangent plane to the cone at P.

Theorem.5. If P(x₁, y₁, z₁) is a point on the cone S = 0, then the equation of the tangent plane to the cone at P is S₁ = 0.

Proof. Given equation to the cone S = S(x, y, z) = 0

Let the equation to the line passing through P(x₁, y₁, z₁) be

⇒ \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}(=r)\) …..(1)

P(x₁, y₁, z₁) ∈ S = 0 ⇒ S₁₁ = 0

The point (lr + x₁, mr + y₁, nr + z₁) of the line (1) lies on S = 0

<=> S(lr + x₁, mr + y₁, nr + z₁) = 0

<=> r²E(l, m, n)⁰ + 2r[lU₁ + mV₁ + nW₁] + S₁₁ < 0

The line is a tangent line to the cone

<=> (lU₁ + mV₁ + nW₁)² – E(l, m, n).S11 = 0

<=> lU₁ + mV₁ + nW₁ = 0 …..(2)

Eliminating l, m, n in (1) and (2), the locus of the tangent line is

(x – x₁)U₁ + (y – y₁)V₁ + (z – z₁)W₁ = 0

i.e. U₁x + V₁y + W₁z = U₁x₁ + V₁y₁ + W₁z₁

i.e. S₁ = S₁₁ i.e. S₁ = 0 [∵ S₁₁ = 0]

∴ The equation to the tangent plane at P(x₁, y₁, z₁) to the cone S = 0 is S₁ = 0

Corollary. If the equation of the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

then the equation to the tangent plane at (x₁, y₁, z₁) on the cone is U₁x + V₁y + W₁z = 0

i.e. x(ax₁ + hy₁ + gz₁ ) + y(hx₁ + by₁ + fz₁) + z[gx₁ + fy₁ + cz₁ ] = 0

i.e. axx₁ + byy₁ + czz₁ + f(y₁z + yz₁) + g(z₁x + zx₁) + h(x₁y + xy₁) = 0

Note 1. The tangent plane at a point P to the cone is also the tangent plane at every point on the generator.

2. The tangent plane at point P to the cone contains the generator through P.

3. The equation to the normal line to the tangent plane at P(x1, y1, z1) is \(\frac{x-x_1}{\mathrm{U}_1}=\frac{y-y_1}{\mathrm{~V}_1}=\frac{z-z_1}{\mathrm{~W}_1}\)

Theorem.6. The necessary and sufficient condition for the plane π = lx + my + nz = 0 to be a tangent plane to the cone E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 is

=\(\left|\begin{array}{lllc}
a & h & g & l \\
h & b & f & m \\
g & f & c & n \\
l & m & n & o
\end{array}\right|=0\)

(1) Necessary Condition

Let P(x₁, y₁, z₁) be the point of contact of the given tangent plane π with the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

∴ The equation to the tangent plane is U₁x + V₁y + W₁z  = 0

⇒ (ax₁ + hy₁ + gz₁)x + (hx₁ + by₁ + fz₁ )y + (gx₁ + fy₁ + cz₁ )z = 0

Comparing with the given tangent plane π i.e. lx + my + nz = 0

We have \(\frac{\mathrm{U}_1}{l}=\frac{\mathrm{V}_1}{m}=\frac{\mathrm{W}_1}{n}\) (= -k, where k ≠ 0)

⇒ \(\frac{a x_1+h y_1+g z_1}{l}=\frac{h x_1+b y_1+f z_1}{m}=\frac{g x_1+f y_1+c z_1}{n}=-k\)

⇒ ax₁ + hy₁ + gz₁ + lk = 0; hx₁ + by₁ + fz₁ + mk = 0; gx₁ + fy₁ + cz₁ + nk = 0

Also lx₁ + my₁ + nz₁ = 0.

The non-zero solution (x₁, y₁, z₁, k) satisfy the equations

ax + hy + gz + lt = 0; hx + by + fz + mt = 0 …..(1)

gx + fy + cz + nt = 0; lx + my + nz = 0.

Hence =\(\left|\begin{array}{lllc}
a & h & g & l \\
h & b & f & m \\
g & f & c & n \\
l & m & n & o
\end{array}\right|=0\)

⇒ \(\Rightarrow p=\mathrm{A} l^2+\mathrm{B} m^2+\mathrm{C} n^2+2 \mathrm{Fmn}+2 \mathrm{G} n l+2 \mathrm{H} l m=0\)

where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h in the determinant.

⇒ \(\Delta=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)

(2) Sufficiency of the Condition. Given ρ = 0, to prove the plane π is a tangent plane to the cone E(x, y, z) = 0.

Proof. If ρ = 0, there exists a non-zero solution (x₁, y₁, z₁, k) to I.

If k = 0, then U₁ = 0, V₁ = 0, W₁ = 0 ⇒ (x₁, y₁, z₁) is the vertex of the cone.

This contradicts the fact that (x₁, y₁, z₁) ≠ (0, 0, 0). Hence k ≠ 0.

Corresponding to the non-zero solution (x₁, y₁, z₁, k) of the equation I, we have

U₁ = -kl, V₁ = -km, W₁ = -kn …..(1) and lx₁ + my₁ + nz₁ = 0 …..(2)

(2) ⇒ P ∈ π

and E(x₁, y₁, z₁) = U₁x₁ + V₁y₁ + W₁z₁ = -k(lx₁ + my₁ + nz₁) = 0.

∴ (x₁, y₁, z₁) is a point on the cone.

∴ P(x₁, y₁, z₁) is a common point of the plane π and the cone.

∴ The equation to the tangent plane at P to the cone is U₁x + V₁y + W₁z  = 0

Since l:m:n = U₁ : V₁ : W₁, lx + my + nz = 0 is the tangent plane P(x₁, y₁, z₁) to the cone.

The Cone Reciprocal Cone

Theorem.7. The locus of the lines perpendicular to the tangent planes of the cone E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 and passing through its vertex is the cone.

⇒ \(\left|\begin{array}{llll}
a & b & g & x \\
h & b & f & y \\
g & f & c & z \\
x & y & z & 0
\end{array}\right|\)

Proof. Let \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) be a line perpendicular to a tangent plane of the cone and passing through the vertex (0, 0, 0).

The cone is E(x, y, z) ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

∴ lx + my + nz = 0 is the tangent plane to (1).

<=> Al² + Bm² + Cn²+ 2Fmn + 2Gnl + 2Hlm = 0

where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h in the determinant.

⇒ \(\Delta=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)

Hence the locus of the normal line is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

i.e.\(\left|\begin{array}{llll}
a & b & g & x \\
h & b & f & y \\
g & f & c & z \\
x & y & z & 0
\end{array}\right|=0\)

This equation represents a cone called the reciprocal cone of (1).

Corollary. The reciprocal cone of

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0 …..(2)

is the cone E(x, y, z) = 0

Proof. By the above theorem, the reciprocal cone of (2) is

A’x² + B’y² + C’z²+ 2F’yz + 2G’zx + 2H’xy = 0 …..(3)

Where A’, B’, C’, F’, G’, H’ are the cofactors of A, B, C, F, G, H in determinant

⇒\(\Delta=\left|\begin{array}{lll}
\mathrm{A} & \mathrm{H} & \mathrm{G} \\
\mathrm{H} & \mathrm{B} & \mathrm{F} \\
\mathrm{G} & \mathrm{F} & \mathrm{C}
\end{array}\right|\)

∴ A’ = BC – F² = (ca – g²)(ab – h²) – (gh – af)²

= a (abc + 2fgh – af² – bg² – ch²) = aΔ,

where Δ = abc + 2fgh – af² – bg² – ch²

Similarly B’ = bΔ, C’ = cΔ, F’ = fΔ, G’ = gΔ, H’ = hΔ

Then the equation (3) reduces to ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

i.e. E(x, y, z) = 0

Thus cones (1) and (2) are Reciprocal cones to each other.

Note 1. A cone and its reciprocal cone will have the same vertex.

2. Corresponding to each tangent plane of a cone there exists a generator of the reciprocal cone which is perpendicular to the tangent plane and vice versa.

3. A cone E(x, y, z) = 0 has three mutually perpendicular tangent planes

<=> the reciprocal cone of E(x, y, z) = 0 has three mutually perpendicular generators

<=> (bc – f²) – (ca – g²) – (ab – h²) = 0

<=> bc + ca + ab = f² + g² + h²

The Cone Solved Problems

Example.1. The semi-vertical angle of a right circular cone having three mutually perpendicular (1) generators is tan^-1√2 (2) tangent plane is \(\tan ^{-1} \frac{1}{\sqrt{2}}\)

Solution. Let the equations to the right circular cone be x² + y² = z² tan²α

(1) If the cone contains three mutually perpendicular generators then a + b + c = 0

(2) The given cone contains three mutually perpendicular tangent planes

<=> its reciprocal cone contains three mutually perpendicular generators

∴ Equations to the reciprocal cone of (1) is

-tan²αx² – tan²αy² + 1.z² = 0    …..(2)

Equation (2) will have three mutually perpendicular generators if

-tan²α – tan²α + 1 = 0 ⇒ \(\alpha=\tan ^{-1} \frac{1}{\sqrt{2}}\).

Example.2. Show that the general equation to a cone that touches the three coordinate planes is \(\sqrt{a x}+\sqrt{b y}+\sqrt{c z}=0\).

Solution. The general equation of the cone containing the three coordinates axes is ayz + bzx + cxy = 0

The reciprocal cone of (1) will have the coordinate planes.

∴ The equation to the reciprocal cone of (1) is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

Where A = -a²; B = -b²; C = -c²

F = bc; G = ca; H = ab

∴ The equation to the required cone is

-a²x² – b²y² – c²z² + 2bcyz + 2cazx + 2abxy = 0

⇒ (ax + by – cz)² = 4abxy ⇒ ax + by – cz = ±2√abxy

⇒ ax + by ± 2√abxy = cz ⇒ (√ax ± √by)²= cz

⇒ √ax ± √by = ±√cz ⇒ √ax ± √by ± √cz = 0

Example.3. Show that the reciprocal cone of ax² + by² + cz² = 0 is the cone \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\).

Solution. Given cone is ax² + by² + cz² = 0

∴ The equation to the reciprocal cone is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

Where A = bc, B = ca, C = ab, F = 0, G = 0, H = 0

∴ The equation to the reciprocal cone is

bcx² + cay² + abz² = 0 ⇒ \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\)

Example.4. Find the equation of the tangent planes to the cone 9x² – 4y² + 16z² = 0, which contains the line \(\frac{x}{32}=\frac{y}{72}=\frac{z}{27}\).

Solution. The given line \(\frac{x}{32}=\frac{y}{72}=\frac{z}{27}\) is the line of intersection of the planes

72x – 32y = 0 and 27y – 72x = 0

i.e., 9x – 4y = 0 and 3y – 8z = 0 …..(1)

∴ The plane passing through line (1) is 9x – 4y + λ(3y – 8z) = 0

i.e., 9x + y(3λ – 4) – 8λz = 0 …..(2)

∴ The equation to the normal line of (2) is\(\frac{x}{9}=\frac{y}{3 \lambda-4}=\frac{z}{-8 \lambda}\) …..(3)

Now plane (2) is a tangent plane to the cone 9x² – 4y² + 16z² = 0 …..(4)

<=> The normal line(3) is a generator of the reciprocal cone of the cone (4)

∴ The equation of the reciprocal cone of (4) is \(\frac{x^2}{9}-\frac{y^2}{4}+\frac{z^2}{16}=0\) …..(5)

Since (3) is a generator of (5)

⇒ \(\frac{9^2}{9}-\frac{(3 \lambda-4)^2}{4}+\frac{(-8 \lambda)^2}{16}=0 \text { i.e. } 7 \lambda^2+24 \lambda+20=0 \Rightarrow \lambda=-2 \text { or } \frac{-10}{7}\)

Hence the equations of tangent planes from (2) are 9x – 10y + 16z = 0 and 63x – 58y + 80z = 0

Example.5. Prove that the equation \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\) represents a cone that touches the coordinate planes and find its reciprocal cone.

Solution. The given equation is \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\)

⇒ \(f x+g y \pm 2 \sqrt{f g x y}=h z\)

⇒ \((f x+g y-h z)^2=4 f g x y\)

⇒ \(f^2 x^2+g^2 y^2+h^2 z^2-2 g h y z-2 h f x x-2 f g x y=0\) …..(1)

This being a homogenous equation of the second degree, represents a quadric cone

The coordinate plane x = 0 meets (1) in \(g^2 y^2+h^2 z^2-2 g h y z=0 \Rightarrow(g y-h z)^2=0\)

⇒ which being a perfect square

⇒ x = 0 touches it similarly we can show that y = 0, z = 0 also touch (1).

Again from the cone(1).

⇒ \(a^{\prime}=f^2, b^{\prime}=g^2, c^{\prime}=h^2, f^{\prime}=-g h,^{\prime} g^{\prime}=-h f,^{\prime} h^{\prime}=-f g\)

∴ \(\mathrm{A}=b c-f^2=g^2 h^2-(-g h)^2=0\)

Similarly \(\mathrm{B}=0, \mathrm{C}=0, \mathrm{~F}=g h-a f=(-h f)(-f g)-f^2(-g h)=2 f^2 g h\)

Similarly \(\mathrm{G}=2 g^2 h f, \mathrm{H}=2 h^2 f g\)

∴ Reciprocal cone is Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

⇒ \(2 f^2 g h y z+2 g^2 h f z x+2 h^2 f g x y=0 \Rightarrow f y z+g z x+h x y=0\)

Example.6. Find the condition that one plane ux + vy + wz = 0 may touch the cone ax² + by² + cz² = 0

Solution. Equation to the normal to the given plane is \(\frac{x}{u}=\frac{y}{v}=\frac{z}{w}\) …..(1)

The equation to the reciprocal cone of

⇒ \(a x^2+b y^2+c z^2=0 \text { is } \frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\) …..(2)

Now the plane touches the cone (2)

<=> The normal of the plane lies on cone (2)

<=> \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=0\) which is the required condition.

Example.7. Find the equation of the cone that touches the three coordinate planes and the planes x + 2y + 3z = 0, 2x + 3y + 4z = 0.

Solution. The equation to the cone touching the three axes can be taken as \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\)

Its reciprocal cone is fyz + gzx + hxy = 0 …..(2)

The planes x + 2y + 3z = 0 and 2x + 3y + 4z = 0

touch the cone (1) <=> their normals lies on (2)

⇒ D.r’s of the normal (1, 2, 3) and (2, 3, 4) satisfy (2)

(1) f(2)(3) + g(3)(1) + h(1)(2) = 0 ⇒ 6f + 3g + 2h = 0 …..(3)

(2) f(3)(4) + g(4)(2) + h(2)(3) = 0 ⇒ 6f + 4g + 3h = 0 …..(4)

Solving (3) and (4): \(\frac{f}{9-8}=\frac{g}{12-18}=\frac{h}{24-18}\)

Hence (1) becomes \(\sqrt{x}+\sqrt{-6 y}+\sqrt{6 z}=0\)

The Cone Intersection Of Two Cones With A Common Vertex

In general two cones with a common vertex intersect along four common generators.

Let S = 0 and S’ = 0 be two cones with origin as the common vertex, then S + λS’ = 0 represents the general equation of a cone whose vertex is at the origin and which passes through the four common generators of the two cones.

Cor. Let λ be so chosen that S + λS’ = 0 becomes the product of two linear factors the two linear factors equated to zero represent the equations to a pair of planes through the common generators.

In that case the values of λ are the roots of the λ – cubic eqution

⇒ \(\left|\begin{array}{lll}
a+\lambda a^{\prime} & h+\lambda h^{\prime} & g+\lambda g^{\prime} \\
h+\lambda h^{\prime} & b+b^{\prime} \lambda & f+\lambda f^{\prime} \\
g+\lambda g^{\prime} & f+\lambda f^{\prime} & c+\lambda c^{\prime}
\end{array}\right|=0\)

The three values of λ give the three pairs of planes through the four common generators.

The Cone Solved Problems

Example.1. Find the equation of the cone which passes through the common generators of the cones 2x² – 4y² – z² = 0 and 10xy – 2yz + 5zx = 0 may the line with direction ratios (1, 2, 3).

Solution.

Given

2x² – 4y² – z² = 0 and 10xy – 2yz + 5zx = 0

Let the required cone be 2x² – 4y² – z² + λ(10xy – 2yz + 5zx) = 0 …..(1)

This is a quadric cone with a vertex at the origin.

The line with d.r.’s (1, 2, 3) lies on (1)

<=> 2(1)² – 4(2)² – (3)² + λ[10(1)(2) – 2(2)(3) + 5(3)(1)] = 0

⇒ – 23 + 23λ = 0 ⇒ λ = 1

∴ Required cone is 2x² – 4y² – z² + 10xy – 2yz + 5zx = 0

Example 2. Find the condition that the lines of the section of the plane lx + my + nz = 0 and the cones ax² + by² + cz² = 0 and fyz + gzx + hxy = 0 should be consistent.

Solution. Any cone through the intersection of the two given cones is ax² + by² + cz² + λ(fyz + gzx + hxy) = 0 …..(1)

Given that the plane lx + my + nz = 0 cuts (1) in coincident lines

⇒ for some value of λ(1) must represent a pair of planes.

Let l₁x + m₁y + n₁z = 0 …..(2) be the other plane.

Then ax² + by² + cz² + λ(fyz + gzx + hxy) = (lx + my + nz)(l1x + m1y + n1z)

⇒ ll₁ = a, mm₁ = b, nn₁ = c

⇒  l₁ = a/1, m₁ = b/m, n₁ = c/n

Again \(\lambda f=m n_1+m_1 n=\frac{c m}{n}+\frac{b n}{m}=\frac{c m^2+b n^2}{m n}\)

Similarly \(\lambda g=\frac{a n^2+c l^2}{n l} \text { and } \lambda h=\frac{a m^2+b l^2}{l m}\)

⇒ \(\frac{c m^2+b n^2}{f m n}=\frac{a n^2+c l^2}{g n l}=\frac{a m^2+b l^2}{h l m}\) which is the required condition.