Let $S$,$S'$ be the foci and $BB'$ be the minor axis of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$ If $\angle BSS' = \theta ,$ then the eccentricity $e$ of the ellipse is equal to

JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Application +4 | -1

Let

S

S'

be the foci and

BB'

be the minor axis of the ellipse

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

\angle BSS' = \theta ,

then the eccentricity

e

of the ellipse is equal to

View answer and explanation

Correct option: B

\cos \theta

The correct answer is

\cos \theta

because it represents the eccentricity of the ellipse, derived from the given angle and the properties of the 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