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UPSC Maths 2013 Paper 1 Q4b — Solution

15 marks · Section A

Question

A cone has for its guiding curve the circle $x^{2}+y^{2}+2ax+2by=0,\,z=0$ and passes through a fixed point $(0,0,c)$. If the section of the cone by the plane $y=0$ is a rectangular hyperbola, prove that the vertex lies on the fixed circle

Technique

Parametric cone construction via projection to base plane; impose two extra conditions to obtain two locus equations for the vertex.

Solution

Strategy. Parameterise the cone by an unknown vertex $V=(\alpha,\beta,\gamma)$ and the given base circle $C_1$ in $z=0$. Impose two conditions: cone passes through $(0,0,c)$, and the section $y=0$ is a rectangular hyperbola. Each gives one equation in $(\alpha,\beta,\gamma)$; together they describe the locus.

Step 1 — Cone equation via projection to base plane

Let $V=(\alpha,\beta,\gamma)$ (with $\gamma\ne 0$ since otherwise the vertex lies in the base plane). A point $(x,y,z)$ lies on the cone iff the line from $V$ through $(x,y,z)$ meets $z=0$ at a point $(X,Y,0)$ on $C_1$.

Parametrise the line as $V+t\,(P-V)$ where $P=(x,y,z)$; setting $z$-component zero gives $t=\gamma/(\gamma-z)$. Then

The base-circle condition $X^{2}+Y^{2}+2aX+2bY=0$ becomes, after multiplying by $(\gamma-z)^{2}$,

Step 2 — Use $(0,0,c)$ on the cone

Substitute $x=0,y=0,z=c$:

Divide by $c$ (assuming $c\ne 0$):

Step 3 — Section by $y=0$ is a rectangular hyperbola

Setting $y=0$ in $(\ast)$ and collecting like terms in the $(x,z)$ plane:

Rectangular hyperbola criterion for $A x^{2}+Bxz+Cz^{2}+\ldots=0$: $A+C=0$ (the asymptotes are perpendicular). Hence

Step 4 — Combine (I) and (II)

Expand (I): $c\alpha^{2}+c\beta^{2}+2c(a\alpha+b\beta)-2\gamma(a\alpha+b\beta)=0$, i.e.

From (II): $\alpha^{2}+\beta^{2}+2a\alpha+2b\beta=-\gamma^{2}$. Substitute:

Since $\gamma\ne 0$,

Step 5 — The locus

Equations (II) and (III) describe the vertex $V=(\alpha,\beta,\gamma)$. Renaming $(\alpha,\beta,\gamma)\to(x,y,z)$:

Answer