JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The foci of the ellipse

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

and the hyperbola

\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}

coincide. Then, the value of

{b^2}

View answer and explanation

Correct option: C

Option C (7) is correct because the value of b² is calculated by equating the foci of the ellipse and hyperbola.

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\frac{x^2}{144}+\frac{y^2}{81}=1

m

\mathrm{M}

\mathrm{mM}

P

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

S

S'.

A

PSS',

A

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

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