JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

Let

P

be a point on the ellipse

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

of eccentricity

e

. If

A,A'

are the vertices and

S,S'

are the foci of the ellipse, then Area

\Delta PSS':\mathrm{Area}\,\,\Delta APA' =

View answer and explanation

Correct option: C

e:1

Option C (

e:1

) is correct because the ratio of the areas of the two triangles is equal to the eccentricity of the ellipse.

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