jee-main 2024 Q66

jee-main · India · session1_31jan_shift1 Conic sections Focal Distance and Point-on-Conic Metric Computation
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
(1) $7 \sqrt { \frac { 2 } { 5 } } - \frac { 8 } { 3 }$
(2) $14 \sqrt { \frac { 2 } { 5 } } - \frac { 4 } { 3 }$
(3) $14 \sqrt { \frac { 2 } { 5 } } - \frac { 16 } { 3 }$
(4) $7 \sqrt { \frac { 2 } { 5 } } + \frac { 8 } { 3 }$ 
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\\
(1) $7 \sqrt { \frac { 2 } { 5 } } - \frac { 8 } { 3 }$\\
(2) $14 \sqrt { \frac { 2 } { 5 } } - \frac { 4 } { 3 }$\\
(3) $14 \sqrt { \frac { 2 } { 5 } } - \frac { 16 } { 3 }$\\
(4) $7 \sqrt { \frac { 2 } { 5 } } + \frac { 8 } { 3 }$
Paper Questions
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