Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{3}$

Answer 1

Answer: (B), (D)

$\frac{5\pi }{3}$

Equation of the tangent to the ellipse from the given point is
$\frac{x}{16}\cdot 4\cos a+\frac{11y}{256}\cdot \frac{16}{\sqrt{11}}\sin a=1\Rightarrow 4x\cos a+\sqrt{11}y\sin a=16$

It is tangent to the circle $x^2+y^2-2x-15=0$
So, the perpendicular distance from centre, i.e., $(1,0)$ to the tangent is equal to the radius of the circle.
$\frac{16-4\cos a}{\sqrt{16{\cos }^{2}a+11{\sin }^{2}a}}=4\Rightarrow (\cos a-4{)}^2=16{\cos }^{2}a+11{\sin }^{2}a\Rightarrow {\cos }^{2}a-8\cos a+16=5{\cos }^{2}a+11\Rightarrow 4{\cos }^{2}a+8\cos a-5=0\Rightarrow \cos a=\frac{1}{2},\frac{-5}{2}\Rightarrow \cos a=\frac{-5}{2}\text{is not possible}\Rightarrow \cos a=\frac{1}{2}\Rightarrow a=\frac{\pi }{3},\frac{5\pi }{3}$