Question

The product of the slopes of the common tangents of the ellipse $x^{2}+4y^{2}=16$ and the parabola $y^{2}-4x-4=0$ is

• A. $-\frac{1}{15}$
• B. $-\frac{1}{5}$
• C. $-\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

Given, ellipse: $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ and parabola: $y^{2}=4\left(x + 1\right)$
Putting $x+1=X$ and $Y=y$ , we get,
$Y^{2}=4X$
So, the equation of the tangent is $Y=mX+\frac{1}{m}=m\left(x + 1\right)+\frac{1}{m}$
$Y=mx+\left(m + \frac{1}{m}\right)\ldots \left(i\right)$
If $\left(i\right)$ is also a tangent to $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ ,
then, $c^{2}=a^{2}m^{2}+b^{2}$
$\Rightarrow \left(m + \frac{1}{m}\right)^{2}=16m^{2}+4$
$\Rightarrow m^{2}+\frac{1}{m^{2}}+2=16m^{2}+4$
$\Rightarrow 15m^{2}-\frac{1}{m^{2}}+2=0$
Let, $m^{2}=t>0$
$\Rightarrow 15t^{2}+2t-1=0$
$\Rightarrow t=\frac{1}{5},\frac{- 1}{3}\Rightarrow m^{2}=\frac{1}{5}$
Hence, $m=\pm\sqrt{\frac{1}{5}}$