JEE MATHEMATICS MCQ: ELLIPSE

1 Medium

1 of 1

Analyzing +4 | -1

P C P^{\prime}

and

D C D^{\prime}

are conjugate diameters of an ellipse

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

and

\theta

is the eccentric angle of

P

. Then, eccentric angle of the point where the circle

P P^{\prime} D

again cuts the ellipse

View answer and explanation

Correct option: D

\frac{\pi}{2}-3 \theta

Option D (

\frac{\pi}{2}-3 \theta

) is correct because it is derived from the property that the sum of eccentric angles where a circle cuts an ellipse is an even multiple of

\pi

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