JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

If the eccentricity of the two ellipse

\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{25}} = 1

and

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

are equal, then the value of

\frac{a}{b}

View answer and explanation

Correct option: C

\frac{{13}}{5}

Option C (

\frac{{13}}{5}

) is correct because it represents the ratio a/b when the eccentricities of the two given ellipses are equal.

More Questions from ELLIPSE

Practice All Chapter MCQs →

\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