First slide

Ellipse

Question

If $l x + m y + n = 0$ is an equation of the line joining the extremities of a pair of semi-conjugate diameters of t the ellipse
$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1,$ then $\frac{9 l^{2} + 4 m^{2}}{n^{2}}$ is equal to

Moderate

A

– 1
B

0
C

1
D

2

Solution

Let the extremities of a pair of semi-conjugate
diameters of the ellipse be $P (3 \cos θ, 2 \sin θ)$ $Q (3 \cos α, 2 \sin α), C$
being the centre.
Then slope of $C P \times$ slope of $C Q = - \frac{4}{9}$
$\Rightarrow \frac{2}{3} \tan θ \times \frac{2}{3} \tan α = - \frac{4}{9}$
$\Rightarrow 1 + \tan θ \tan α = 0 \Rightarrow α - θ = π / 2$
$\Rightarrow$ coordinates of $Q$ are $(- 3 \sin θ, 2 \cos θ)$
equation of $P Q$ is
$2 (\sin θ - \cos θ) x - 3 (\cos θ + \sin θ) y + 6 = 0$
which represents $l x + m y + n = 0$ , then
$\begin{aligned} \frac{l}{2 (\sin θ - \cos θ)} = \frac{m}{- 3 (\cos θ + \sin θ)} = \frac{n}{6} \\ So \frac{9 l^{2} + 4 m^{2}}{n^{2}} = (\sin θ - \cos θ)^{2} + (\cos θ + \sin θ)^{2} = 2 . \end{aligned}$

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