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  • If latus rectum of the ellipse <img alt="\mathrm{x^2 \tan ^2 \alpha+y^2 \sec ^2 \alpha=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2%20%5Ctan%20%5E2%20%5Calpha&plus

If latus rectum of the ellipse \mathrm{x^2 \tan ^2 \alpha+y^2 \sec ^2 \alpha=1} is \frac{1}{2} then \alpha(0<\alpha<\frac{\pi }{3}) is equal to
 

Option: 1

\frac{\pi }{12}


Option: 2

\frac{\pi }{6}


Option: 3

5\frac{\pi }{12}


Option: 4

none of these


Answers (1)

best_answer

\begin{aligned} &\mathrm{L.R. =\frac{2 \cdot \cos ^2 \alpha}{\cot \alpha}=\frac{1}{2}}\\ &\mathrm{so \sin 2 \alpha=\frac{1}{2} ~or ~2 \alpha=\frac{\pi}{6}, \frac{5 \pi}{6}}\\ &\mathrm{\therefore \quad \alpha=\frac{\pi}{12}, ~as ~0<\alpha<\frac{\pi}{3}q}\\ \end{aligned}
Hence (a) is the correct answer.

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Riya

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