Question

If the point of intersection of the ellipses $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 and $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}$ = 1 be at the extremities of the conjugate diameters of the former, then-

• A. $\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 2$
• B. $\frac{a^{2}}{\alpha^{2}}–\frac{b^{2}}{\beta^{2}} = 2$
• C. $\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 0$
• D. $\frac{a^{2}}{\alpha^{2}}–\frac{b^{2}}{\beta^{2}} = 0$

Answer 1

Answer: (A)

$\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 2$

Answer 2

The points of intersection of the ellipses $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 and $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}$= 1 are given by

x² $\left( \frac{1}{\alpha^{2}} - \frac{1}{a^{2}} \right)$ + y² $\left( \frac{1}{\beta^{2}} - \frac{1}{b^{2}} \right)$ = 0.

This equation represents a pair of straight lines passing through the origin.

Obviously here, they correspond to the diameters PCP¢ and DCD¢.

Let y = mx be one of the lines. Then y must satisfy the equation

$\left( \frac{1}{\alpha^{2}} - \frac{1}{a^{2}} \right)$ + m²$\left( \frac{1}{\beta^{2}} - \frac{1}{b^{2}} \right)$ = 0.

This is a quadratic equation in ‘m’ and hence it will have two roots, say m₁ and m₂.

\ m₁m₂ = $\left( \frac{1}{\alpha^{2}} - \frac{1}{a^{2}} \right)$ ø$\left( \frac{1}{b^{2}} - \frac{1}{\beta^{2}} \right)$.

If the lines PCP¢ and DCD¢ are conjugate diameters, then we must have$\frac{\frac{1}{\alpha^{2}} - \frac{1}{a^{2}}}{\frac{1}{\beta^{2}} - \frac{1}{b^{2}}}$= – $\frac{b^{2}}{a^{2}}$

Ž $\frac{a^{2}}{\alpha^{2}}$ – 1 = –$\left( \frac{b^{2}}{\beta^{2}} - 1 \right)$ = 1 – $\frac{b^{2}}{\beta^{2}}$

\ $\frac{a^{2}}{\alpha^{2}}$+$\frac{b^{2}}{\beta^{2}}$ = 2.

Hence (1) is the correct answer.