Maximum length of chord of ellipse frac{{{x^2}}}{8} + frac{{{y^2}}}{4} = 1 , such that ec

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10-2. Parabola, Ellipse, Hyperbola

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

A

$4$

B

$2\sqrt 2 $

C

$16$

D

$8$

Solution

$a=2 \sqrt{2}, b=2$

Let $P \equiv(2 \sqrt{2} \cos \theta, 2 \sin \theta)$

and $Q \equiv(-2 \sqrt{2} \sin \theta, 2 \cos \theta)$

$\therefore(P Q)^{2}=8(\cos \theta+\sin \theta)^{2}+4(\sin \theta-\cos \theta)^{2}$

$=12+4 \sin 2 \theta \leq 16$

$(P Q)_{\max }=4$

Maximum value of sin function is $1 .$

Standard 11

Mathematics

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