Maximum length of chord of the ellipse x2-8-y2-4-1- such that eccentric angles of its extremities differ by -2 is

If the line $x + 2 y + 4 = 0$ cutting the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in points whose eccentric angles are $30^{\circ}$ and $60^{\circ}$ subtends a right angle at the origin then its equation is

Q. The equation of chord of ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

whose sum and difference of eccentric angles are

\frac{π}{3}

and

\frac{2 π}{3}

respectively is

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Maximum length of chord of the ellipse x28+y24=1, such that eccentric angles of its extremities differ by π2 is

The correct option is D 4a=2√2,b=2Let P≡(2√2cosθ,2sinθ)and Q≡(−2√2sinθ,2cosθ)∴(PQ)2=8(cosθ+sinθ)2+4(sinθ−cosθ)2=12+4sin2θ≤16(PQ)max=4Maximum value of sin function is 1.

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 $\frac{π}{2}$ is

The correct option is D $4$
$a = 2 \sqrt{2}, b = 2$
Let $P \equiv (2 \sqrt{2} cos θ, 2 sin θ)$
and $Q \equiv (- 2 \sqrt{2} sin θ, 2 cos θ)$
$∴ (P Q)^{2} = 8 (cos θ + sin θ)^{2} + 4 (sin θ - cos θ)^{2}$
$= 12 + 4 sin 2 θ \leq 16$
$(P Q)_{m a x} = 4$
Maximum value of sin function is $1.$