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
$y=mx+c$
Since, $y=mx+c$ is a tangent to Eq. (i).
$\therefore c^2=a^2{m}^2+b^2$
$\Rightarrow c^2=49m^2+4$ ...(iii)
Similarly, $y=mx+c$ 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
$49m^2+4=16\left(1+m^2\right)$
$\Rightarrow 49m^2-16m^2=16-4$
$\Rightarrow 33m^2=12$
$\Rightarrow m^2=\frac{4}{11}$
$\Rightarrow m=\frac{2}{\sqrt{11}}$