JEE MATHEMATICS MCQ: 2D GEOMETRY SYSTEM

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Application +4 | -1

Two points

A

and

B

have coordinates

(1,0)

and

(-1,0)

respectively and

Q

is a point which satisfies the relation

A Q-B Q= \pm 1

. The locus of

Q

View answer and explanation

Correct option: B

12 x^{2}-4 y^{2}=3

Option B (

12 x^{2}-4 y^{2}=3

) is correct because the given condition

AQ - BQ = \pm 1

defines a hyperbola, and after simplification, the equation becomes

12x^2 - 4y^2 = 3

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)