The maximum length of a chord of the ellipse | vedclass

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(A) For the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,we have $a^2 = 8$ and $b^2 = 4$,so $a = 2\sqrt{2}$ and $b = 2$.Let the eccentric angles of the

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$4$

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(A) For the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,we have $a^2 = 8$ and $b^2 = 4$,so $a = 2\sqrt{2}$ and $b = 2$.Let the eccentric angles of the extremities of the chord be $	heta$ and $	heta + \frac{\pi}{2}$.The coordinates of the points are $P = (2\sqrt{2} \cos 	heta, 2 \sin 	heta)$ and $Q = (2\sqrt{2} \cos(	heta + \frac{\pi}{2}), 2 \sin(	heta + \frac{\pi}{2})) = (-2\sqrt{2} \sin 	heta, 2 \cos 	heta)$.The square of the length of the chord $PQ$ is given by $L^2 = (2\sqrt{2} \cos 	heta + 2\sqrt{2} \sin 	heta)^2 + (2 \sin 	heta - 2 \cos 	heta)^2$.$L^2 = 8(\cos 	heta + \sin 	heta)^2 + 4(\sin 	heta - \cos 	heta)^2$.$L^2 = 8(1 + \sin 2	heta) + 4(1 - \sin 2	heta) = 12 + 4 \sin 2	heta$.To maximize $L$,we set $\sin 2	heta = 1$,which gives $L^2 = 12 + 4 = 16$.Thus,the maximum length $L = \sqrt{16} = 4$.

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

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