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{xx_{1}}{a^{2}} -\frac{yy_{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 \quad $[$\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 }$