JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the foci of a hyperbola are same as that of the ellipse \(\frac{x^2}{9}+\frac{y^2}{25}=1\) and the eccentricity of the hyperbola is \(\frac{15}{8}\) times the eccentricity of the ellipse, then the smaller focal distance of the point \(\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)\) on the hyperbola, is equal to
- A \(7 \sqrt{\frac{2}{5}}-\frac{8}{3}\)
- B \(14 \sqrt{\frac{2}{5}}-\frac{4}{3}\)
- C \(14 \sqrt{\frac{2}{5}}-\frac{16}{3}\)
- D \(7 \sqrt{\frac{2}{5}}+\frac{8}{3}\)
Answer & Solution
Correct Answer
(A) \(7 \sqrt{\frac{2}{5}}-\frac{8}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{9}+\frac{y^2}{25}=1\) \(a=3, b=5 \) \(e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm b e)=(0, \pm 4)\) \(\therefore e_H=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}\) Let equation hyperbola…
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