Question

If lx +my + n = 0 represents a chord of the ellipse

b²x² + a²y² = a²b² whose eccentric angles differ by 90⁰, then:

• A. a²l² + b²m² = n²
• B. $\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} - b^{2})^{2}}{n^{2}}$
• C. a²l² + b²m² = 2n²
• D. None of these

Answer 1

Answer: (C)

a²l² + b²m² = 2n²

Answer 2

Equation of chord joining points P (a cos α, b sin α) and Q (a cos β, b sin β) is

$\frac{x}{a}\cos\left( \frac{\alpha + \beta}{2} \right) + \frac{y}{b}\sin\left( \frac{\alpha + \beta}{2} \right) = \cos\left( \frac{\alpha - \beta}{2} \right)$Now

β = α + 90⁰

$\frac{x}{a}\cos\left( \frac{2\alpha + 90^{0}}{2} \right) + \frac{y}{b}\sin\left( \frac{2\alpha + 90^{0}}{2} \right) = \frac{1}{\sqrt{2}}$

Now compare it with lx + my = -n

$\frac{\cos\left( \frac{2\theta + 90^{0}}{2} \right)}{al} = \frac{\sin\left( \frac{2\alpha + 90^{0}}{2} \right)}{bm} = - \frac{1}{\sqrt{2}n}$

⇒ cos²θ + sin²θ = 1

⇒ a²l² + b²m² = 2n².