Question

The slope of a common tangent to the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{4}=1$ and the circle $x^{2}+y^{2}=16$ is

• A. $\frac{5}{\sqrt{11}}$
• B. $\frac{4}{\sqrt{11}}$
• C. $\frac{3}{\sqrt{11}}$
• D. $\frac{2}{\sqrt{11}}$

Answer 1

Answer: (D)

Given equation are
$\frac{x^{2}}{49}+\frac{y^{2}}{4}=1$ ...(i)
and $x^{2}+y^{2}=16$ ...(ii)
Let equation of common tangent be
$y=m x+ c$
Since, $y=m x +c$ is a tangent to Eq. (i).
$\therefore c^{2}=a^{2} m^{2}+b^{2}$
$\Rightarrow c^{2}=49 m^{2}+4$ ...(iii)
Similarly, $y=m x+ c$ is a tangent to Eq. (ii)
$\therefore c^{2}=a^{2}\left(1+m^{2}\right)$
$\Rightarrow c^{2}=16\left(1+m^{2}\right)$ ...(iv)
From Eqs. (iii) and (iv), we get
$49 m^{2}+4=16\left(1+m^{2}\right)$
$\Rightarrow 49 m^{2}-16 m^{2}=16-4$
$\Rightarrow 33 m^{2}=12$
$\Rightarrow m^{2}=\frac{4}{11}$
$\Rightarrow m=\frac{2}{\sqrt{11}}$