JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

If the ellipse

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

and the circle

x^2 + y^2 = r^2

where

b < r < a

intersect in four points and the slope of a common tangent to the ellipse and the circle is

\frac{b}{a},

then

r^2

is equal to

View answer and explanation

Correct option: B

\frac{2a^2b^2}{a^2+b^2}

Option B (

\frac{{2a^2b^2}}{{a^2+b^2}}

) is correct because it is derived by finding the condition for a common tangent to the ellipse and the 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