UPSC 2018 Maths Optional Paper 1 Q4c — Step-by-Step Solution

13 marks · Section A

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

Question

Find the equation of the cone with $(0,0,1)$ as the vertex and $2x^2-y^2=4,\ z=0$ as the guiding curve.

Technique

Generator method — join vertex to guiding curve, parametrise, eliminate the parameter using the plane $z=0$ relation $t=1-z$ .

Solution

Step 1 — Generator from vertex to a base point

A cone is the union of all lines (generators) joining the vertex $V=(0,0,1)$ to points of the guiding curve in $z=0$ . Let $(x,y,z)$ be a general point of the cone and let the generator through it meet $z=0$ at $(X,Y,0)$ , which must satisfy $2X^2-Y^2=4$ .

Step 2 — Parametrise the generator and eliminate the parameter

Points on the line $V\to(X,Y,0)$ are $(x,y,z)=(0,0,1)+t(X,Y,-1)=(tX,\ tY,\ 1-t)$ . From the $z$ -coordinate, $t=1-z$ . Hence (for $z\neq1$ )

X=\frac{x}{1-z},\qquad Y=\frac{y}{1-z}.

Step 3 — Substitute into the guiding-curve relation

Impose $2X^2-Y^2=4$ :

2\Big(\frac{x}{1-z}\Big)^2-\Big(\frac{y}{1-z}\Big)^2=4\ \Longrightarrow\ \frac{2x^2-y^2}{(1-z)^2}=4.

Therefore

Answer

\boxed{\;2x^2-y^2=4(1-z)^2,\quad\text{i.e.}\quad 2x^2-y^2-4z^2+8z-4=0.\;}