Question

If the tangent at the point $\left(4 \cos \theta, \frac{16}{\sqrt{11}} \sin \theta\right)$ to the ellipse $16 x^{2}+11 y^{2}=256$ is also a tangent to the circle $x ^{2}+ y ^{2}-2 x =15$, then the value of $\theta$ is

• A. $\pm \frac{\pi}{4}$
• B. $\pm \frac{\pi}{3}$
• C. $\pm \frac{\pi}{6}$
• D. $\pm \frac{\pi}{2}$

Answer 1

Answer: (B)

The equation of the tangent at $\left(4 \cos \theta, \frac{16}{\sqrt{11}} \sin \theta\right)$ to
the ellipse $16 x^{2}+11 y^{2}=256$ is
$16 x(4 \cos \theta)+11 y\left(\frac{16}{\sqrt{11}} \sin \theta\right)=256$
or $4 x \cos \theta+\sqrt{11} y \sin \theta=16$
This touches the circle $( x -1)^{2}+ y ^{2}=4^{2}$ if
$\left|\frac{4 \cos \theta-16}{\sqrt{16 \cos ^{2} \theta+11 \sin ^{2} \theta}}\right|=4 $
$\Rightarrow (\cos \theta-4)^{2}=16 \cos ^{2} \theta+11 \sin ^{2} \theta$
$\Rightarrow 15 \cos ^{2} \theta+11 \sin ^{2} \theta+8 \cos \theta-16=0$
$\Rightarrow 4 \cos ^{2} \theta+8 \cos \theta-5=0$
$\Rightarrow (2 \cos \theta-1)(2 \cos \theta+5)=0$
$\Rightarrow \cos \theta=\frac{1}{2} $
$\Rightarrow \theta=\pm \frac{\pi}{3}$