Question

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

• A. 9
• B. 4
• C. 5
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Ellipse 5x² + 9y² = 45

Ž $\frac{x^{2}}{9} + \frac{y^{2}}{5}$ = 1

e = $\sqrt{1 - \frac{5}{9}}$ = $\frac{2}{3}$

Hyperbola 5x² – 4y² = 45

$\frac{x^{2}}{9}$ – $\frac{y^{2}}{\frac{45}{4}}$ = 1

Ž e´ = $\sqrt{1 + \frac{45/4}{9}}$ = $\frac{3}{2}$Žee´ = 1