Question

A straight line touches the rectangular hyperbola $9x^2-9y^2=8$ and the parabola $y^2=32x$. An equation of the line is:

• 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)

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

Let the equation be $y=mx+c$.

Since it touches the hyperbola
$9x^2-9y^2=8$ or $\frac{x^2}{8/9}-\frac{y^2}{8/9}=1$
(with $a^2=b^2=8/9$), we get

$c^2=\left(8/9\right)(m^2-1)$ (i)

Also, since it touches the parabola $y^2=32x$, we get

$c=8/m...\left(ii\right)$
From (i) and (ii) we get,
$\frac{8}{9}(m^2-1)=\frac{64}{m^2}$ $\Rightarrow$ $m^2(m^2-1)=72$

$\Rightarrow$ $m^4-m^2-72=0$ $\Rightarrow$ $(m^2-9)(m^2+8)=0$
$\Rightarrow$ $m=\pm 3$ [$m^2=-8$ is rejected]

From (ii), $c=8/3$ and -8/3 when $m=3$ and -3 respectively. Therefore, the required equations are

$y=3x+\frac{8}{3}$ and $y=-3x-\frac{8}{3}$
or $9x-3y+8=0$ or $9x+3y+8=0$