JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

The eccentricity of an ellipse is

\frac{1}{3}

, latus rectum is 4 and centre is

(0,0)

, then the equation of the ellipse is (given major axis is along

x

-axis)

View answer and explanation

Correct option: B

\frac{16 x^2}{81}+\frac{2 y^2}{9}=1

Option A (

\frac{16 x^2}{81}+\frac{2 y^2}{9}=1

) is correct because it satisfies the given eccentricity and latus rectum conditions for the ellipse.

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