$AB$ is a diameter of ${x^2} + 9{y^2} = 25.$ The eccentric angle of $A$ is $\pi /6$. Then, the eccentric angle of $B$ is

JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

AB

is a diameter of

{x^2} + 9{y^2} = 25.

The eccentric angle of

A

\pi /6

. Then, the eccentric angle of

B

View answer and explanation

Correct option: B

- 5\,\pi /6

Option B (

\ - 5\\,\pi /6

) is correct because it represents the eccentric angle of point B, which is diametrically opposite to point A.

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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