---
title: "Condition for second degree equation to be a cone"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:condition-second-degree-equation-be"
source: https://www.bctengineer.com/q/semester_2/mathematics/condition-second-degree-equation-be
site: BCT Engineering
---

# Condition for second degree equation to be a cone

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2072 Ashwin, 5 marks; 2068 Magh, 5 marks (asked 2 times)
- Video: [Prove that given equation is a cone | #analytical Solid Geometry | SEM-6 | UNIT-2](https://www.youtube.com/watch?v=abR52IuJawU) — VSP UNITY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/condition-second-degree-equation-be
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Prove that the equation ax²+by²+cz²+2ux+2vy+2wz+d = 0 represents a cone if u²/a + v²/b + w²/c = d.

## Short answer

If u²/a + v²/b + w²/c = d, the equation becomes a(x+u/a)² + b(y+v/b)² + c(z+w/c)² = 0, a homogeneous second-degree equation representing a cone with vertex (−u/a, −v/b, −w/c).

## Answer

A cone with vertex (x₀, y₀, z₀) has an equation that is homogeneous of the second degree in (x − x₀), (y − y₀), (z − z₀). So we show that under the given condition the equation can be written in that form.

#### Complete the squares
Write the surface as

```text
a x² + 2ux + b y² + 2vy + c z² + 2wz + d = 0
a(x² + 2ux/a) + b(y² + 2vy/b) + c(z² + 2wz/c) + d = 0
a(x + u/a)² − u²/a + b(y + v/b)² − v²/b + c(z + w/c)² − w²/c + d = 0
```

So

```text
a(x + u/a)² + b(y + v/b)² + c(z + w/c)² + [ d − (u²/a + v²/b + w²/c) ] = 0
```

#### Apply the condition
If u²/a + v²/b + w²/c = d, the bracket vanishes and the equation becomes

```text
a(x + u/a)² + b(y + v/b)² + c(z + w/c)² = 0
```

Shift the origin to the point V(−u/a, −v/b, −w/c), putting X = x + u/a, Y = y + v/b, Z = z + w/c:

aX² + bY² + cZ² = 0.

This is homogeneous of degree 2 in X, Y, Z. If (X, Y, Z) satisfies it, so does (kX, kY, kZ) for every k. So every point on the surface lies on a line through V that lies wholly on the surface: the surface is a cone with vertex V(−u/a, −v/b, −w/c). ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What form must the equation of a cone with vertex (x0, y0, z0) take?
   A. Homogeneous of second degree in (x - x0), (y - y0), (z - z0)
   B. Linear in x, y, z
   C. Homogeneous of first degree in x, y, z
   D. Any second-degree equation
   Answer: A. Scaling a point's displacement from the vertex must keep it on the surface. (Key point: What defines a cone)

2. After completing squares, what is the constant term?
   A. d + (u^2/a + v^2/b + w^2/c)
   B. d - (u^2/a + v^2/b + w^2/c)
   C. d - (u/a + v/b + w/c)
   D. d
   Answer: B. Each square a(x + u/a)^2 subtracts u^2/a. (Key point: Completing the squares)

3. What happens when u^2/a + v^2/b + w^2/c = d?
   A. The equation becomes linear
   B. All coefficients become zero
   C. The constant term vanishes, leaving a(x + u/a)^2 + b(y + v/b)^2 + c(z + w/c)^2 = 0
   D. The equation becomes a sphere
   Answer: C. With the bracket zero, the equation is a sum of squared terms equal to 0. (Key point: Using the condition)

4. What is the vertex?
   A. (u/a, v/b, w/c)
   B. (-u, -v, -w)
   C. (-a/u, -b/v, -c/w)
   D. (-u/a, -v/b, -w/c)
   Answer: D. Shifting the origin to X = x + u/a etc. makes aX^2 + bY^2 + cZ^2 = 0 homogeneous, with vertex at X = Y = Z = 0. (Key point: Vertex)

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