Question

$P:y^2=8x,E:\frac{x^2}{4}+\frac{y^2}{15}=1$

Equation of a tangent common to both the parabola $P$ and the ellipse $E$ is

• A. $x-2y+8=0$
• B. $x+2y+8=0$
• C. $x+2y-8=0$
• D. $x-2y-8=0$

Answer 1

Answer: (A), (B)

The correct options are
A $x-2y+8=0$
B $x+2y+8=0$

Equation of tangent to the parabola $y^2=8x$ is given by,
$y=mx+\frac{2}{m}..\left(1\right)$
And the equation of tangent to the ellipse $\frac{x^2}{4}+\frac{y^2}{15}=1$. is given by,
$y=mx+\sqrt{4m^2+15}..\left(2\right)$
Now since both (1) and (2) are same,
$\Rightarrow \frac{4}{m^2}=4m^2+15\Rightarrow 4m^4+15m^2-4=0$
$\Rightarrow m=\pm \frac{1}{2}$
Thus required equation is given by,
$x-2y+8=0$ or $x+2y+8=0$