UPSC 2014 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

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

Question

Examine whether the plane $x+y+z=0$ cuts the cone $yz+zx+xy=0$ in perpendicular lines.

Technique

Direct substitution (real-roots check) + standard cone-plane perpendicularity formula.

Solution

Strategy. Two routes:

Direct algebra: parametrise the plane, substitute into the cone, check if the resulting equation has two real (perpendicular) lines.
Standard formula: for a cone $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy=0$ cut by a plane $lx+my+nz=0$ , the two lines are perpendicular iff

(b+c)l^{2}+(c+a)m^{2}+(a+b)n^{2}-2fmn-2gnl-2hlm=0.

We use both routes for cross-checking.

Step 1 — Direct algebra

Eliminate $z=-(x+y)$ from $yz+zx+xy=0$ :

y[-(x+y)]+[-(x+y)]x+xy=-xy-y^{2}-x^{2}-xy+xy=-(x^{2}+xy+y^{2}).

The locus on the plane is $x^{2}+xy+y^{2}=0$ .

Over $\mathbb R$ : $x^{2}+xy+y^{2}=\left(x+\tfrac{y}{2}\right)^{2}+\tfrac{3}{4}y^{2}\ge 0$ , with equality iff $x=y=0$ . So only the origin satisfies the system — no real lines of intersection.

Step 2 — Standard perpendicularity formula

For our cone $yz+zx+xy=0$ : $a=b=c=0$ , $2f=2g=2h=1$ , so $f=g=h=\tfrac{1}{2}$ .

Plane: $l=m=n=1$ .

Compute:

(b+c)l^{2}+(c+a)m^{2}+(a+b)n^{2}-2fmn-2gnl-2hlm=0+0+0-1-1-1=-3.

Non-zero. So the perpendicularity condition fails.

Conclusion

Answer

\boxed{\;\text{The plane does not cut the cone in perpendicular lines (and in fact does not meet it in any real lines beyond the origin).}\;}