JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

If the normal at the end of latusrectum of the ellipse

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

passes through

\left( {0,\; - b} \right)

, then

{e^4} + {e^2}

(where

e

is eccentricity) equals

View answer and explanation

Correct option: A

The correct answer is 1 because the given condition implies a relationship between the eccentricity and the semi-major/minor axes, leading to

{e^4} + {e^2} = 1

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