Suppose $S$ and $S'$ are foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1.$ If $P$ is a variable point on the ellipse and if $\Delta $ is area of the triangles $PSS'$, then the maximum value $\Delta $ is

JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

Suppose

S

and

S'

are foci of the ellipse

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1.

P

is a variable point on the ellipse and if

\Delta

is area of the triangles

PSS'

, then the maximum value

\Delta

View answer and explanation

Correct option: B

Option B (12) is correct because it represents the maximum area of the triangle formed by the foci and a point on the 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