First slide

Ellipse

Question

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

Moderate

Solution

$\begin{array}{l} Let P \equiv (2 \sqrt{2} \cos θ, 2 \sin θ) \\ Q \equiv (- 2 \sqrt{2} \sin θ, 2 \cos θ) \\ (P Q)^{2} = 8 (\cos θ + \sin θ)^{2} + 4 (\sin θ - \cos θ)^{2} \end{array}$
$\begin{array}{l} = 12 + 4 \sin 2 θ \leq 16 \\ ∴ {((P Q)^{2})}_{max} = 16 \end{array}$

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