JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The perpendicular from the centre of an ellipse

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

to polar of a point with respect to the ellipse is constant equal to

c

. Then, the locus of the point is the ellipse

View answer and explanation

Correct option: C

\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c^2}

Option C (

\frac{{x^2}}{{a^4}} + \frac{{y^2}}{{b^4}} = \frac{{1}}{{c^2}}

) is correct because it represents the locus of the point when the perpendicular distance from the center to the polar is constant.

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