JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

If the eccentricities of the ellipse

\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1

and the hyperbola

\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{{b^2}}} = 1

are reciprocals of each other, then

{b^2}

is equal to

View answer and explanation

Correct option: A

192

Option A (192) is correct because the eccentricities are reciprocals, leading to the equation 1 + b²/64 = 4, which solves to b² = 192.

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\mathrm{M}

\mathrm{mM}

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\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

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r

\frac{x^2}{25}+\frac{y^2}{16}=1

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O