If the tangent to the ellipse $x^{2} + 4y^{2} = 16$ at the point $P( heta)$ is a normal to circle $

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If the tangent to the ellipse $x^{2} + 4y^{2} = 16$ at the point $P(	heta)$ is a normal to circle $x^{2} + y^{2} - 8x - 4y = 0$ then $	heta$ equals:

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Sol. $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$Equation of tangent at $P(	heta)$ is $\frac{xcos	heta}{4} + \frac{ysin	heta}{2} = 1$tangent passes through $(4,2)$$${\Rightarrow cos	heta + sin	heta = 1 }{\Rightarrow sin\left( 	heta + \frac{\pi}{4} ight) = \frac{1}{\sqrt{2}} }{\Rightarrow \ 	heta + \frac{\pi}{4} = \frac{\pi}{4},\frac{3\pi}{4} }{\Rightarrow \ 	heta = 0,\frac{\pi}{2}}$$

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