JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Understanding +4 | -1

A point

P

moves so that sum of its distances from

\left( {-ae,0} \right)

and

\left( {ae,0} \right)

2a

. Then, the locus of

P

View answer and explanation

Correct option: D

\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{a^2}\left( {1 + {e^2}} \right)}} = 1

Option B (

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2}\left( {1 - {e^2}} \right)}} = 1

) is correct because the sum of distances from two points is constant, defining an ellipse.

More Questions from ELLIPSE

Practice All Chapter MCQs →

\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