UPSC Maths 2025 Paper 1 Q1e — Solution

10 marks · Section A

Question

Find the equation of the cone whose vertex is the point $(1, 1, 0)$ and whose guiding curve is $y = 0$ , $x^2 + z^2 = 4$ .

Technique

Treat the cone as the locus of generators (lines) through the vertex meeting the guiding curve: parametrise the line from the vertex, find where it meets the plane $y=0$ , and substitute into the curve equation.

Solution

Step 1 — Generator through vertex and a general point.

Let $P=(X,Y,Z)$ be a general point of the cone. The generator is the line through the vertex $V=(1,1,0)$ and $P$ : $(x,y,z) = (1,1,0) + t\big(X-1,\; Y-1,\; Z-0\big).$

Step 2 — Meet the plane of the guiding curve $y=0$ .

Set the $y$ -coordinate to $0$ : $1 + t(Y-1) = 0 \;\Rightarrow\; t = \frac{1}{1-Y}\quad (Y\ne1).$

At this $t$ , the $x$ - and $z$ -coordinates of the intersection point are $x^* = 1 + t(X-1) = 1 + \frac{X-1}{1-Y} = \frac{(1-Y)+(X-1)}{1-Y} = \frac{X-Y}{1-Y},$ $z^* = t\,Z = \frac{Z}{1-Y}.$

Step 3 — Impose the guiding curve $x^{*2}+z^{*2}=4$ .

$\left(\frac{X-Y}{1-Y}\right)^2 + \left(\frac{Z}{1-Y}\right)^2 = 4.$ Multiply through by $(1-Y)^2$ : $(X-Y)^2 + Z^2 = 4(1-Y)^2.$

Step 4 — Expand and simplify (rename $X,Y,Z\to x,y,z$ ): $x^2 - 2xy + y^2 + z^2 = 4(1 - 2y + y^2)$ $x^2 - 2xy + y^2 + z^2 = 4 - 8y + 4y^2$ $x^2 - 2xy - 3y^2 + z^2 + 8y - 4 = 0.$

Answer

$\boxed{\;x^2 - 2xy - 3y^2 + z^2 + 8y - 4 = 0.\;}$