If normal to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(a e, \frac{b^2}{a}\right)$ is passing through $(0,-2 b)$, then value of eccentricity is

JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

If normal to ellipse

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

\left(a e, \frac{b^2}{a}\right)

is passing through

(0,-2 b)

, then value of eccentricity is

View answer and explanation

Correct option: C

\sqrt{2(\sqrt{2}-1)}

Option C,

\sqrt{2(\sqrt{2}-1)}

, is the correct eccentricity value derived by using the equation of the normal to the ellipse and the given point.

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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