Question

If pairs of lines $3x^2-2pxy-3y^2=0$ and $5x^2-2qxy-5y^2=0$ are such that each pair bisects the angle between the other pair, then $pq$ is equal to

• A. $-1$
• B. $-3$
• C. $-5$
• D. $-15$

Answer 1

The correct option is D $-15$

Given pairs

$3x^2-2pxy-3y^2=0$----(1)

$5x^2-2qxy-5y^2=0$----(2)

Equation (1) can be written as

$3(x^2-y^2)=2pxy$----(3)

Eq of angle bisector of eq (2)

$\frac{x^2-y^2}{5+5}=\frac{xy}{-q}$

$\frac{x^2-y^2}{10}=\frac{xy}{-q}$----(4)

Now as the given in question the eq of angle bisector of one pair bisects the other pair

So dividing eq (4) by (3)

$\frac{\frac{x^2-y^2}{10}}{3(x^2-y^2)}=\frac{\frac{xy}{-q}}{2pxy}$

$\frac{1}{30}=\frac{1}{-2pq}$

$pq=-15$