Question

If e and e¢ are the eccentricities of the ellipse 5x² + 9y² = 45 and the hyperbola 5x² –4y² = 45 respectively, then ee¢ is equal to

• A. 1
• B. 4
• C. 5
• D. 9

Answer 1

Answer: (A)

1

Answer 2

5x² + 9y² = 45 ….(1)

$\frac{x^{2}}{9}$+ $\frac{y^{2}}{5}$= 1 ̃ e =$\sqrt{1 - \frac{b^{2}}{a^{2}}}$=$\sqrt{1 - \frac{5}{9}}$=$\frac{2}{3}$

& 5x² – 4y² = 45

$\frac{x^{2}}{9}$–$\frac{y^{2}}{(45/4)}$= 1 ̃ e¢ =$\sqrt{1 + \frac{b^{2}}{a^{2}}}$

=$\sqrt{1 + \frac{45}{4 \times 9}}$=$\sqrt{\frac{81}{4 \times 9}}$= $\frac{2}{3}$

\ ee' = 1