JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

P C P^{\prime}

and

D C D^{\prime}

form a pair of conjugate diameters of the ellipse

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

and

R

is any point on the circle

x^2+y^2=c^2

, then

View answer and explanation

Correct option: D

P R^2+D R^2+P^{\prime} R^2+D^{\prime} R^2=2\left(a^2+b^2+2 c^2\right)

Option D (

P R^2 + D R^2 + P' R^2 + D' R^2 = 2(a^2 + b^2 + 2 c^2)

) is correct because it represents the sum of squared distances from points on the ellipse and circle.

More Questions from ELLIPSE

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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