JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

1 of 1

Application +4 | -1

Locus of all such points so that sum of its distances from

(2,-3)

and

(2,5)

is always 10, is

View answer and explanation

Correct option: D

\frac{(x-2)^2}{9}+\frac{(y-1)^2}{25}=1

Option D (

\frac{{(x-2)^2}}{{9}}+\frac{{(y-1)^2}}{{25}}=1

) is correct because the locus is an ellipse with foci (2,-3) and (2,5), major axis 10, and center (2,1).

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