First slide

Ellipse

Question

If a, b are the eccentric angles of the
extremities of a focal chord of the ellipse
$x^{2} / 16 + y^{2} / 9 = 1$ then $\tan (α / 2) \tan (β / 2) =$

Easy

A

$\frac{\sqrt{7} + 4}{\sqrt{7} - 4}$
B

$- \frac{9}{23}$
C

$\frac{\sqrt{5} - 4}{\sqrt{5} + 4}$
D

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

Solution

The eccentricity $e = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$
Let $P (4 \cos α, 3 \sin α)$ and $Q (4 \cos β, 3 \sin β)$
be a focal chord of the ellipse passing
through the focus at $(\sqrt{7}, 0)$
Then $\frac{3 \sin β}{4 \cos β - \sqrt{7}} = \frac{3 \sin α}{4 \cos α - \sqrt{7}}$
$\begin{array}{ll} \Rightarrow \frac{\sin (α - β)}{\sin α - \sin β} = \frac{\sqrt{7}}{4} \\ \Rightarrow \frac{\cos [(α - β) / 2]}{\cos [(α + β) / 2]} = \frac{\sqrt{7}}{4} \\ \Rightarrow \tan (\frac{α}{2}) \tan (\frac{β}{2}) = \frac{\sqrt{7} - 4}{\sqrt{7} + 4} = \frac{23 - 8 \sqrt{7}}{- 9} \end{array}$

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