Question

If the tangent at the point $\left( 4\cos\varphi,\frac{16}{\sqrt{11}}\sin\varphi \right)$to the ellipse 16x² + 11y² = 256 is also a tangent to the circle

x² + y² – 2x = 15, then the value of f is -

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

Answer 1

Answer: (C)

± $\frac{\pi}{3}$

Answer 2

The equation of the tangent at $\left( 4\cos\varphi,\frac{16}{\sqrt{11}}\sin\varphi \right)$

to the ellipse 16x² + 11y² = 256 is

16x (4 cos f) + 11y $\left( \frac{16}{\sqrt{11}}\sin\varphi \right)$ = 256

or 4x cos f +$\sqrt{11}$y sin f = 16

This touches the circle (x – 1)² + y² = 4², therefore

$\left| \frac{4\cos\varphi - 16}{\sqrt{16\cos^{2}\varphi + 11\sin^{2}\varphi}} \right|$= 4

Ž (cos f – 4)² = 16 cos² f + 11 sin² f

Ž 15cos² f + 11 sin² f + 8 cos f – 16 = 0

Ž 4 cos² f + 8 cos f – 5 = 0

Ž (2 cos f – 1) (2 cos f + 5) = 0 Ž cos f = $\frac{1}{2}$ Ž f = ± $\frac{\pi}{3}$.