Question

The equation of common tangent of hyperbola $9x^2-9y^2=8$ and $y^2=32x$ 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 $9x+3y+8=0$

Given equation of hyperbola is $9x^2-9y^2=8$
$\Rightarrow \frac{x^2}{8/9}-\frac{y^2}{8/9}=1$
and equation of parabola is $y^2=32x$.
Let the equation of tangent is $y=mx+c\dots \left(1\right)$
and $c^2=\frac{8}{9}(m^2-1)\dots \left(2\right)$
Line $\left(1\right)$ is tangent to the parabola So,
$c=\frac{8}{m}\dots \left(3\right)$
From equation $\left(2\right)$ and $\left(3\right)$
$\frac{8^2}{m^2}=\frac{8}{9}(m^2-1)$
$\Rightarrow m^4-m^2-72=0$
$\Rightarrow m^2=-8,9$
$\Rightarrow m=\pm 3,c=\pm \frac{8}{3}$
By equation $\left(1\right)$ and $\left(3\right)$ common tangents will be
$3y-9x-8=0,3y+9x+8=0$