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Question: If latus rectum of the ellipse \(x^{2}\tan^{2}\alpha + y^{2}\sec^{2}\alpha = 1\)is \(\frac{1}{2}\) t...

If latus rectum of the ellipse x2tan⁡2α+y2sec⁡2α=1x^{2}\tan^{2}\alpha + y^{2}\sec^{2}\alpha = 1is 12\frac{1}{2} then α(0<α<π)\alpha(0 < \alpha < \pi) is equal to

A

π/12

B

π/6

C

5π/12

D

None of these

Answer

5π/12

Explanation

Solution

x2tan⁡2α+y2sec⁡2α=1x^{2}\tan^{2}\alpha + y^{2}\sec^{2}\alpha = 1

⇒x2cot⁡2α+y2cos⁡2α=1\frac{x^{2}}{\cot^{2}\alpha} + \frac{y^{2}}{\cos^{2}\alpha} = 1

∵cos⁡2α=cot⁡2α(1−e2)\because\cos^{2}\alpha = \cot^{2}\alpha\left( 1 - e^{2} \right)

⇒sin⁡2α=1−e2,e2=cos⁡2α\sin^{2}\alpha = 1 - e^{2},e^{2} = \cos^{2}\alpha

∴e=cos⁡α\therefore e = \cos \alpha (α≠900)\left( \alpha \neq 90^{0} \right)

∵\becauselatus rectum = 12=2b2a\frac{1}{2} = \frac{2b^{2}}{a}

⇒ a=4b2a = 4b^{2}

⇒ cot⁡α=4cos⁡2α\cot\alpha = 4\cos^{2}\alpha

⇒1sin⁡α=4cos⁡α\frac{1}{\sin\alpha} = 4\cos\alpha

⇒sin⁡2α=12\sin 2\alpha = \frac{1}{2}

⇒2α=nπ+(−1)nπ62\alpha = n\pi + ( - 1)^{n}\frac{\pi}{6}

⇒α=nπ2+(−1)nπ12\alpha = \frac{n\pi}{2} + ( - 1)^{n}\frac{\pi}{12} for n = 1

α=π2−π12=5π12\alpha = \frac{\pi}{2} - \frac{\pi}{12} = \frac{5\pi}{12}