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 $\tan (α / 2) \tan (β / 2) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

$- \frac{9}{23}$

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed 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 $\frac{3 \sin β}{4 \cos β - \sqrt{7}} = \frac{3 \sin α}{4 \cos α - \sqrt{7}}$

$\begin{array}{ll} \Rightarrow \frac{\sin (α - β)}{\sin α - \sin β} = \frac{\sqrt{7}}{4} \\ \Rightarrow \frac{\cos [(α - β) / 2]}{\cos [(α + β) / 2]} = \frac{\sqrt{7}}{4} \\ \Rightarrow \tan (\frac{α}{2}) \tan (\frac{β}{2}) = \frac{\sqrt{7} - 4}{\sqrt{7} + 4} = \frac{23 - 8 \sqrt{7}}{- 9} \end{array}$