Question

The condition on a and b for which two distinct chords of the ellipse $\frac{x^{2}}{2a^{2}} + \frac{y^{2}}{2b^{2}}$ = 1, passing through (a, –b) are bisected by the line x + y = b, is –

• A. a² + 6ab £ 7b²
• B. a² + 6ab ³ 7b²
• C. a² + ab £ 7b²
• D. a² + ab ³ 7b²

Answer 1

Answer: (B)

a² + 6ab ³ 7b²

Answer 2

Let (a, b – a) be a point on the line x + y = b such that the chord of the given ellipse passing through (a, –b) are bisected at (a, b – a).

Then, the equation of the chord is

$\frac{\alpha x}{2a^{2}} + \frac{(b - \alpha)y}{2b^{2}}$= $\frac{\alpha^{2}}{2a^{2}}$+ $\frac{(b - \alpha)^{2}}{2b^{2}}$

[Using : S¢ = T]

This passes through (a, –b). Therefore,

$\frac{\alpha a}{2a^{2}}–\frac{(b - \alpha)b}{2b^{2}}$= $\frac{\alpha^{2}}{2a^{2}}$+ $\frac{(b - \alpha)^{2}}{2b^{2}}$

Ž a² (a² + b²) – ab (3a + b) a + 2a²b² = 0

Since a is real. Therefore,a²b²(3a + b)² – 8a²b² (a² + b²) ³ 0

Ž a² + 6ab – 7b² ³ 0

Ž a² + 6ab ³ 7b², which is the required condition.