JEE MATHEMATICS MCQ: ELLIPSE

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P

is any point lying on the ellipse

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

, whose foci are

S

and

S^{\prime}

. Let

\angle P S S^{\prime}=\alpha

and

\angle P S^{\prime} S=\beta

, then which of the following is/are true?

View answer and explanation

Correct option: D

\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{\sqrt{a^2-b^2}}{b^2}\left(a-\sqrt{a^2-b^2}\right)

Option D (

\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{\sqrt{{a^2-b^2}}}{b^2}\left(a-\sqrt{{a^2-b^2}}\right)

) is correct because it is derived using the properties of the ellipse and trigonometric relationships involving the angles to the foci.

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