Question

The condition for two diameters of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$represented by Ax² + 2Hxy + By² = 0 to be conjugate is

• A. Ab² = Ba²
• B. Aa² = -Bb²
• C. Aa² = Bb²
• D. None of these

Answer 1

Answer: (C)

Aa² = Bb²

Answer 2

Let the two diameters represented by

Ax² + 2Hxy + By² = 0.

be y = mx and y = m'x.

$\because$ The two diameters are conjugate,

∴ mm' = $\frac{b^{2}}{a^{2}}$ ... (1)

Also m, m' are the roots of Bm² + 2Hm + A = 0

∴ mm' = A/B ... (2)

From (1) and (2), $\frac{b^{2}}{a^{2}} = \frac{A}{B}$ or Aa² = Bb² which is the required condition.