JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

The equation of the ellipse whose foci are at

\left( { \pm \,2,0} \right)

and eccentricity is

\frac{1}{2}

, is

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

, Then,

View answer and explanation

Correct option: A

{a^2} = 16,{b^2} = 12

Option A (

{a^2} = 16,{b^2} = 12

) is correct because it correctly calculates the values of a² and b² based on the given foci and eccentricity.

More Questions from ELLIPSE

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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