UPSC 2017 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Sphere · Analytic Geometry · asked 17× in 13 yrs · Read the full method →

Question

A plane passes through a fixed point $(a,b,c)$ and cuts the axes at the points $A,B,C$ respectively. Find the locus of the centre of the sphere which passes through the origin $O$ and $A,B,C$ .

Technique

Intercept-form plane $+$ origin-passing sphere ( $d=0$ ); read centre $=(p/2,q/2,r/2)$ , then eliminate $p,q,r$ using the fixed-point condition.

Solution

Step 1 — Parametrize the plane by its intercepts

Let the plane meet the axes at $A=(p,0,0),\ B=(0,q,0),\ C=(0,0,r)$ . In intercept form

\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1.

It passes through the fixed point $(a,b,c)$ , so

\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=1.\tag{$\ast$}

Step 2 — Sphere through $O,A,B,C$

A general sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ . Through $O=(0,0,0)$ forces $d=0$ . Then:

A=(p,0,0):\ p^2+2up=0\Rightarrow u=-\tfrac p2;\quad B:\ v=-\tfrac q2;\quad C:\ w=-\tfrac r2.

So the sphere is $x^2+y^2+z^2-px-qy-rz=0$ , with centre

\Big(\tfrac p2,\tfrac q2,\tfrac r2\Big).

Step 3 — Eliminate the parameters $p,q,r$

Let the centre be $(X,Y,Z)$ . Then

X=\tfrac p2,\ Y=\tfrac q2,\ Z=\tfrac r2\ \Longrightarrow\ p=2X,\ q=2Y,\ r=2Z.

Substitute into the constraint $(\ast)$ :

\frac{a}{2X}+\frac{b}{2Y}+\frac{c}{2Z}=1\ \Longrightarrow\ \frac{a}{X}+\frac{b}{Y}+\frac{c}{Z}=2.

Step 4 — State the locus

Dropping to running coordinates $(x,y,z)$ for the centre:

Answer

\boxed{\;\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2.\;}

UPSC 2017 Maths Optional Paper 1 Q2b — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Parametrize the plane by its intercepts

Step 2 — Sphere through O,A,B,CO,A,B,CO,A,B,C

Step 3 — Eliminate the parameters p,q,rp,q,rp,q,r

Step 4 — State the locus

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 2 — Sphere through $O,A,B,C$

Step 3 — Eliminate the parameters $p,q,r$