JEE MATHEMATICS MCQ: 2D GEOMETRY SYSTEM

1 Hard

1 of 1

Application +4 | -1

A point moves such that the sum of its distances from two fixed points

(a e, 0)

and

(-a e, 0)

is always

2 a

. Then equation of its locus is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the equation of an ellipse where the sum of distances from two foci is constant.

More Questions from 2D GEOMETRY SYSTEM

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\alpha

(\alpha, 2 \alpha)

\Delta A B C

A(0,2), B(2,0)

C(4,4)

P

2 P A=3 P B

A

(0,0)

B

(4,-3)

(1, \sqrt{3}),(0,0)

(2,0)

(1,5),(-3,1) \,\&\,(-2,1)

9 x^2-16 y^2-18 x+32 y-23

=0

25 x^2+9 y^2-50 x-18 y+33=0

(0,0)(3,0)

(0,4)