Let e _ 1 be the eccentricity of the hyperbola x ^ 2 16 - y ^ 2 9 =...

The given equation of hyperbola is, $H: \dfrac{{x}^{2}}{16} - \dfrac{{y}^{2}}{9} = 1$

$⇒ {e}_{1} = \sqrt{1 + \dfrac{{3}^{2}}{{4}^{2}}} = \sqrt{\dfrac{16 + 9}{16}}$

$⇒ {e}_{1} = \dfrac{5}{4}$

Foci of this hyperbola is given by, $F\left(\pm ae, 0\right).</p><p>$⇒ F\left(\pm 5, 0\right)$</p><p>It is given that,${e}_{1}{e}_{2} = 1$</p><p>$⇒ {e}_{2} = \dfrac{4}{5}$</p><p>Also, ellipse is passing through$( \pm 5, 0 )$</p><p>$⇒ \dfrac{{5}^{2}}{{a}^{2}} + \dfrac{0}{{b}^{2}} = 1$</p><p>$⇒ \dfrac{25}{{a}^{2}} = 1$</p><p>$⇒ {a}^{2} = 25$</p><p>$⇒ a = 5$</p><p>Now,${e}_{2} = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}}$</p><p>$⇒ \dfrac{4}{5} = \sqrt{1 - \dfrac{{b}^{2}}{25}}$</p><p>$⇒ \dfrac{16}{25} = 1 - \dfrac{{b}^{2}}{25}$</p><p>$⇒ \dfrac{{b}^{2}}{25} = \dfrac{9}{25}$</p><p>$⇒ {b}^{2} = 9$</p><p>$∴ a = 5 and b = 3$</p><p>$E: \dfrac{{x}^{2}}{25} + \dfrac{{y}^{2}}{9} = 1$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5747592f-aef4-40b8-a436-dde2db425e3f.png" style="width: 250px; height: 166px;"/></p><p>Putting,$y = 2$</p><p>$⇒ \dfrac{{x}^{2}}{25} + \dfrac{4}{9} = 1$</p><p>$⇒ \dfrac{{x}^{2}}{25} = \dfrac{5}{9}$</p><p>$⇒ {x}^{2} = \dfrac{125}{9}$</p><p>$⇒ x = \pm \dfrac{5\sqrt{5}}{3}$</p><p>So, end points of chord are \left(\pm \dfrac{5\sqrt{5}}{3}, 2\right)$

Thus, length of chord $PQ$- is given by,

${L}_{PQ} = 2 \times \dfrac{5\sqrt{5}}{3}$

$⇒ {L}_{PQ} = \dfrac{10\sqrt{5}}{3}$