Question

The lines $2x+3y+4=0$ and $3x-2y+5=0$ may be conjugate w.r.t. the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• A. $a^2+b^2=\frac{10}{3}$
• B. $a^2-b^2=\frac{10}{3}$
• C. $b^2+a^2=\frac{10}{3}$
• D. none of these.

Answer 1

Answer: (A)

Let $\left(x_1,y_1\right)$ be the pole of
$2x+3y+4=0$ w.r.t $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Its polar is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$
Also polar is $2x+3y+4=0$
$\therefore \frac{\frac{x_1}{a^2}}{2}=-\frac{\frac{y_1}{b^2}}{3}=-\frac{1}{4}$
$x_1=-\frac{a^2}{2},y_1=-\frac{3b^2}{4}$
This lies on $3x-2y+5=0$ if
$\frac{-3a^2}{2}-\frac{6b^2}{4}+5=0$[$\because$ lines are conjugate]
$\Rightarrow -6a^2-6b^2+20=0$
$\Rightarrow 6\left(a^2+b^2\right)=20$
$\Rightarrow a^2+b^2=\frac{10}{3}$