JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

In the normal at the end of latusrectum of the ellipse

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

with eccentricity

e

, passes through one end of the minor axis, then

View answer and explanation

Correct option: B

{e^2}\left( {1 + {e^2}} \right) = 1

The correct answer is B. By finding the equation of the normal at the end of the latus rectum and using the given condition, the relationship

{e^2}(1 + {e^2}) = 1

is derived.

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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