JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

\alpha

and

\beta

are eccentric angles of the ends of

a

focal chord of the ellipse

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

, then

\tan \frac{\alpha}{2} \tan \frac{\beta}{2}

is equal to

View answer and explanation

Correct option: B

\frac{e-1}{e+1}

Option B (

\frac{e-1}{e+1}

) is correct because it represents the relationship between the tangents of half angles and the eccentricity.

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m

\mathrm{M}

\mathrm{mM}

P

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

S

S'.

A

PSS',

A

A

B

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(-\sqrt{7}, 0)

P

9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

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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