UPSC 2014 Maths Optional Paper 1 Q4a-ii — Step-by-Step Solution

10 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

Prove that the equation $ax^{2}+by^{2}+cz^{2}+2ux+2vy+2wz+d=0$ represents a cone if $\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}=d$ .

Complete the square in each variable to centre the quadratic; cone condition = constant term vanishes.

Strategy. Complete the square in each variable; the resulting form is a cone iff the constant term vanishes.

For each variable, group quadratic and linear terms:

ax^{2}+2ux=a\!\left(x+\dfrac{u}{a}\right)^{2}-\dfrac{u^{2}}{a},

by^{2}+2vy=b\!\left(y+\dfrac{v}{b}\right)^{2}-\dfrac{v^{2}}{b},

cz^{2}+2wz=c\!\left(z+\dfrac{w}{c}\right)^{2}-\dfrac{w^{2}}{c}.

Substituting back into the original equation:

a\!\left(x+\dfrac{u}{a}\right)^{2}+b\!\left(y+\dfrac{v}{b}\right)^{2}+c\!\left(z+\dfrac{w}{c}\right)^{2}-\left(\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}\right)+d=0.

Let $X=x+u/a$ , $Y=y+v/b$ , $Z=z+w/c$ . The equation becomes

aX^{2}+bY^{2}+cZ^{2}=\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}-d.

A surface $aX^{2}+bY^{2}+cZ^{2}=k$ (with $a,b,c$ same sign and $k$ a real constant) is:

So our equation represents a cone iff the constant on the RHS vanishes:

\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}-d=0\;\Longleftrightarrow\;\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}=d.

When this holds, the equation reduces to

aX^{2}+bY^{2}+cZ^{2}=0,

a cone with vertex at $(X,Y,Z)=(0,0,0)$ , i.e., $(x,y,z)=(-u/a,-v/b,-w/c)$ in the original coordinates.

\boxed{\;ax^{2}+by^{2}+cz^{2}+2ux+2vy+2wz+d=0\text{ is a cone iff }\dfrac{u^{2}}{a}+\dfrac{v^{2}}{b}+\dfrac{w^{2}}{c}=d.\;}