Question 18. Let e_{1} be the eccentricity of the hyperbola \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 and e_{2} be the eccentricity of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b, which passes through the foci of the hyperbola. If e_{1} e_{2}=1, then the length of the chord of the ellipse parallel to the x -axis and passing through is :
(1) 4 \sqrt{5}
(2) \frac{8 \sqrt{5}}{3}
(3) \frac{10 \sqrt{5}}{3}
(4) 3 \sqrt{5}
Answer (3)
Explanation:
H: \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \quad e_{1}=\frac{5}{4} \therefore e_{1} e_{2}=1 \Rightarrow e_{2}=\frac{4}{5}Also, ellipse is passing through ( \pm 5,0)
\therefore a=5 and b=3
E: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1
End point of chord are \left( \pm \frac{5 \sqrt{5}}{3}, 2\right)
\therefore L_{PQ}=\frac{10 \sqrt{5}}{3}