JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

If the product of the perpendiculars from the foci of the ellipse

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

upon the polar of

P

is always

c^2

, then locus of

P

is the ellipse

View answer and explanation

Correct option: A

b^4\left(c^2+a^2-b^2\right) x^2+c^2 a^4 y^2=a^4 b^4

Option A (

b^4\left(c^2+a^2-b^2\right) x^2+c^2 a^4 y^2=a^4 b^4

) is correct because it represents the locus of P derived from the product of perpendiculars from the foci to the polar of P being equal to

c^2

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