UPSC 2016 Maths Optional Paper 1 Q4a — Step-by-Step Solution

10 marks · Section A

Second-degree equations in three variables · Analytic Geometry · asked 3× in 13 yrs · Read the full method →

Question

Find the surface generated by a line which intersects the lines $y=a=z,\ x+3z=a=y+z$ and parallel to the plane $x+y=0$ .

Technique

Parametrize a generic generator by its meeting points on the two guide lines; impose the parallel-to-plane condition to cut the family to one parameter; eliminate parameters for the surface equation.

Solution

Step 1 — Parametrize the two guide lines

$L_1:\ y=a,\ z=a$ — a line parallel to the $x$ -axis. Point on $L_1$ : $(t,\,a,\,a)$ .

$L_2:\ x+3z=a,\ y+z=a$ . Put $z=s$ : then $y=a-s$ , $x=a-3s$ . Point on $L_2$ : $(a-3s,\ a-s,\ s)$ .

Step 2 — Direction of the generator and the parallelism condition

A generator joins $(t,a,a)$ on $L_1$ to $(a-3s,a-s,s)$ on $L_2$ ; its direction is

\mathbf d=(a-3s-t,\ -s,\ s-a).

Parallel to the plane $x+y=0$ (normal $(1,1,0)$ ) means $\mathbf d\cdot(1,1,0)=0$ :

(a-3s-t)+(-s)=0\ \Rightarrow\ t=a-4s.

Then $a-3s-t=a-3s-(a-4s)=s$ , so $\mathbf d=(s,\,-s,\,s-a)$ and the $L_1$ end is $(a-4s,\,a,\,a)$ .

Step 3 — Eliminate the parameters to get the surface

A point on the generator: $(X,Y,Z)=(a-4s+su,\ a-su,\ a+(s-a)u)$ for parameters $s,u$ . Eliminating $s,u$ (resultant elimination) gives the single relation

X Y+X Z+Y^2+Y Z-2aX-2aZ=0.

Replacing $(X,Y,Z)\to(x,y,z)$ :

Answer

\boxed{\;y^2+xy+yz+zx-2a(x+z)=0.\;}