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  • If   <img alt="(\sqrt{3}) \mathrm{bx}+\mathrm{ay}=2 \mathrm{ab}" src="https://entrancecorner.oncodecogs.com/gif.latex?%28%5Csqrt%7B3%7D%29%20%5Cmathrm%7Bbx%7D&plus;%5Cmathrm%7Bay%7D%3

If   (\sqrt{3}) \mathrm{bx}+\mathrm{ay}=2 \mathrm{ab}   touches the ellipse  \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1   then eccentric angle of the point of contact is 

Option: 1

\frac{\pi}{6}


Option: 2

\frac{\pi}{4}


Option: 3

\frac{\pi}{3}


Option: 4

\frac{\pi}{2}


Answers (1)

best_answer

Equation of tangent is   \frac{\mathrm{x}}{\mathrm{a}} \frac{\sqrt{3}}{2}+\frac{\mathrm{y}}{\mathrm{b}} \frac{1}{2}=1  and equation of tangent at the point 

(\mathrm{a} \cos \phi, \mathrm{b} \sin \phi) is

\frac{\mathrm{x}}{\mathrm{a}} \cos \phi+\frac{\mathrm{y}}{\mathrm{b}} \sin \phi=1 .

Both are same i.e., \cos \phi=\frac{\sqrt{3}}{2}, \sin \phi=\frac{1}{2}

\Rightarrow \quad \phi=\frac{\pi}{6}

Hence (1) is the correct answer.

Posted by

SANGALDEEP SINGH

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