Question

If a, b are the eccentric angles of the extremities of a focal chord of the ellipse x²/16 + y²/9 = 1, then tan (a/2) tan (b/2) =

• A. $\frac{\sqrt{7} + 4}{\sqrt{7} - 4}$
• B. $- \frac{9}{23}$
• C. $\frac{\sqrt{5} - 4}{\sqrt{5} + 4}$
• D. $\frac{8\sqrt{7} - 23}{9}$

Answer 1

Answer: (D)

$\frac{8\sqrt{7} - 23}{9}$

Answer 2

The equation of the ellipse is of the form
x²/a² + y²/b² = 1 where a² = 16, b² = 9

\ the eccentricity e = $\sqrt{1 - \frac{9}{16}}$=$\frac{\sqrt{7}}{4}$.

Let P(4 cos a, 3 sin a) and Q (4 cos b, 3 sin b) be a focal chord of the ellipse passing through the focus at ($\sqrt{7}$, 0).

Then $\frac{3\sin\beta}{4\cos\beta - \sqrt{7}}$=$\frac{3\sin\alpha}{4\cos\alpha - \sqrt{7}}$

Ž $\frac{\sin(\alpha - \beta)}{\sin\alpha - \sin\beta}$=$\frac{\sqrt{7}}{4}$

Ž $\frac{\cos\lbrack(\alpha - \beta)/2\rbrack}{\cos\lbrack(\alpha + \beta)/2\rbrack}$ = $\frac{\sqrt{7}}{4}$

Ž tan $\left( \frac{\alpha}{2} \right)$tan $\left( \frac{\beta}{2} \right)$ = $\frac{\sqrt{7} - 4}{\sqrt{7} + 4}$= $\frac{23 - 8\sqrt{7}}{- 9}$.