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 (\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}) on the hyperbola, is equal to [31-Jan-2024 Shift 1]

Correct Answer is : 7√25−83

7 \sqrt{\frac{2}{5}}-\frac{8}{3}

14 \sqrt{\frac{2}{5}}-\frac{4}{3}

14 \sqrt{\frac{2}{5}}-\frac{16}{3}

7 \sqrt{\frac{2}{5}}+\frac{8}{3}

JEE Main 31-Jan-2024 Shift 1 Solved Paper

Section: Mathematics

Question : 3 of 90

Marks: +1, -0

If the foci of a hyperbola are same as that of the ellipse ‌

+‌

=1 and the eccentricity of the hyperbola is ‌

times the eccentricity of the ellipse, then the smaller focal distance of the point (√2,‌

√‌

) on the hyperbola, is equal to

[31-Jan-2024 Shift 1]

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