Question

A straight line touches the rectangular hyperbola $9x^2-9y^2=8$ and the parabola $y^2=32x$. The equation of the line(s) is/are :

• A. $9x+3y-8=0$
• B. $9x-3y+8=0$
• C. $9x+3y+8=0$
• D. $9x-3y-8=0$

Answer 1

Answer: (B), (C)

The correct options are
B $9x-3y+8=0$
C $9x+3y+8=0$

The equation of tangent to the given hyperbola at any point $(a\sec \theta ,b\tan \theta )$ is given by
$\frac{x}{a}\sec \theta -\frac{y}{a}\tan \theta =1$, where $a=b=\frac{2\sqrt{2}}{3}$
$\Rightarrow x\sec \theta -y\tan \theta =\frac{2\sqrt{2}}{3}$ ...... $\left(1\right)$
Similarly, the equation of tangent at any point $(8t^2,16t)$ of the parabola $y^2=32x$ is
$ty=x+8t^2$
$\Rightarrow x-ty=-8t^2$ ...... $\left(2\right)$
Comparing equations $\left(1\right)$ and $\left(2\right)$, we get
$\frac{1}{\sec \theta }=\frac{+t}{\tan \theta }=\frac{-8t^2}{2\sqrt{2}}\times 3$ ...... ($\left(1\right)$ & $\left(2\right)$ are same)
$\Rightarrow t=\sin \theta$ ..... $\left(3\right)$
$\Rightarrow cos\theta =-2\sqrt{2}\times 3t^2$ ..... $\left(4\right)$
From (3) and (4) $72t^4+t^2-1=0$
$\Rightarrow t^2=\frac{1}{9}\Rightarrow t=\pm \frac{1}{3}$
Using the above value in equation (2), we get the equation as
$\pm 3y=9x+8$
Hence, the equation of line are

$9x+3y+8=0$ and $9x-3y+8=0$