Q97 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
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$\frac{3}{\sqrt{11}}$
- Correct answerD$\frac{2}{\sqrt{11}}$
Solution
Given equation are
$$
\begin{aligned}
& \frac{x^2}{49}+\frac{y^2}{4}=1 \\
& x^2+y^2=16
\end{aligned}
$$
and
$$
x^2+y^2=16
$$
Let equation of common tangent be
$$
y=m x+c
$$
Since, $y=m x+c$ is a tangent to Eq. (i).
$$
\begin{array}{ll}
\therefore & c^2=a^2 m^2+b^2 \\
\Rightarrow & c^2=49 m^2+4
\end{array}
$$
Similarly, $y=m x+c$ is a tangent to Eq. (ii)
$$
\begin{array}{llrl}
& \therefore & & c^2=a^2\left(1+m^2\right) \\
& \Rightarrow & c^2=16\left(1+m^2\right)
\end{array}
$$
From Eqs. (iii) and (iv), we get
$$
\begin{aligned}
& 49 m^2+4=16\left(1+m^2\right) \\
& \Rightarrow \quad 49 m^2-16 m^2=16-4 \\
& \Rightarrow \quad 33 m^2=12 \\
& \Rightarrow \quad m^2=\frac{4}{11} \Rightarrow m=\frac{2}{\sqrt{11}} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'D'.