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  • Maximum length of chord of the ellipse <img alt="\mathrm{\frac{x^2}{8}+\frac{y^2}{4}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B8%7D&plus;%5Cfrac%

Maximum length of chord of the ellipse \mathrm{\frac{x^2}{8}+\frac{y^2}{4}=1}, such that eccentric angles of its extremities differ by \mathrm{\frac{\pi}{2}} is

Option: 1

3


Option: 2

5


Option: 3

4


Option: 4

8


Answers (1)

best_answer

\mathrm{\therefore \quad P \equiv(2 \sqrt{2} \cos \theta, 2 \sin \theta)}

\mathrm{\begin{aligned} & \text { and } Q \equiv\left(2 \sqrt{2} \cos \left(\frac{\pi}{2}+\theta\right), 2 \sin \left(\frac{\pi}{2}+\theta\right)\right) \\ & \Rightarrow \quad Q \equiv(-2 \sqrt{2} \sin \theta, 2 \cos \theta) \\ & \therefore \quad(P Q)^2=8(\cos \theta+\sin \theta)^2+4(\sin \theta-\cos \theta)^2=12+4 \sin 2 \theta \\ & \therefore \quad(P Q)_{\max .}=4 \end{aligned}}

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