Question
If the tangent at the point $\left( 4cos\theta,\frac{16sin\theta}{\sqrt{11}} \right)$ to ellipse $16x^{2} + 11y^{2} = 256$ is also tangent to circle $x^{2} + y^{2} - 2x = 15$, then $\theta$ equals :
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Solution
Sol. Equation of tangent is
$$\frac{x}{4}cos\theta + \frac{\sqrt{11}ysin\theta}{16} = 1 $$$4cos\theta x + \sqrt{11}sin\theta y = 16\ $ which touches the circle $(x - 1)^{2} + y^{2} = 16$
$${\Rightarrow \frac{|4cos\theta - 16|}{\sqrt{16\cos^{2}\theta + 11\sin^{2}\theta}} = 4 }{\Rightarrow (cos\theta - 4)^{2} = 16\cos^{2}\theta + 11\sin^{2}\theta}$$
$$= \cos^{2}\theta - 8cos\theta + 16$$
$$\Rightarrow 4\cos^{2}\theta + 8cos\theta - 5 = 0$$
$$= 4\cos^{2}\theta - 2cos\theta + 10cos\theta - 5$$
$${\Rightarrow \ (2cos\theta + 5)(2cos\theta - 1) = 0 }{\Rightarrow cos\theta = \frac{1}{2}}$$
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