Q. The slopes of the lines represented by $x^{2} + 2hxy + 2y^{2} = 0$ are in the ratio 1 : 2, then $h$ equals

A) $\pm \frac{1}{2}$
B) $\pm \frac{3}{2}$
C) $\pm 1$
D) $\pm 3$
Explanation

(b)
On comparing the given equation with
${ax}^{2} + 2hxy + {by}^{2} = 0$, we get
$a = 1,\ 2h = 2h$ and $b = 2$
Let the slope of the lines are $m_{1}$ and $m_{2}$.
$\therefore m_{1}:m_{2} = 1:2$
Let $m_{1} = m$ and $m_{2} = 2m$
$\therefore m_{1} + m_{2} = - \frac{2h}{2} \Rightarrow m + 2m = - h\ \Rightarrow \ h = - 3m$ …(i)
and $m_{1}m_{2} = \frac{a}{b}\ \Rightarrow \ m \bullet 2m = \frac{1}{2} \Rightarrow m = \pm \frac{1}{2}$ …(ii)
From Eqs. (i) and (ii), we get
$h = \pm \frac{3}{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.