JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

B

is an end of the minor axis of the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, F_1

and

F_2

are the foci such that the triangle

BF_1F_2

is equilateral, then eccentricity of the ellipse is:

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{{1}}{{2}}

) is correct because the eccentricity is derived from the equilateral triangle condition.

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

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\mathrm{M}

\mathrm{mM}

P

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

S

S'.

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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