Question

If $x^2-2pxy-y^2=0$ and $x^2-2qxy-y^2=0$ bisect angles between each other, then

• A. $p+q=1$
• B. $pq=1$
• C. $pq+1$
• D. $p^2+pq+q^2=0$

Answer 1

Answer: (C)

$pq+1$

The equation of the bisectors of the angles between the lines $x^2-2pxy-y^2=0$

$\frac{x^2-y^2}{1-(-1)}=\frac{xy}{-p}$

$\Rightarrow$ $\frac{x^2-y^2}{2}=\frac{-xy}{p}$

$\Rightarrow$ $p{x}^2+2xy-p{y}^2=0$

This is same as $x^2-2qxy-y^2=0$. Therefore

$\frac{p}{1}=\frac{2}{-2q}=\frac{p}{1}$

$\frac{1}{-q}=\frac{p}{1}\Rightarrow -1=-pq$

$\Rightarrow$ $pq+1=0$