JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

Coordinates of the vertices

B

and

C

of the base of a triangle

A B C

are

(-a, 0)

and

(a, 0)

respectively. If

C-B=\pi / 3

, the vertex

A

lies on the curve

View answer and explanation

Correct option: C

\sqrt{3}\left(x^2-y^2\right)+2 x y-\sqrt{3} a^2=0

Option C (

\sqrt{3}(x^2-y^2)+2 x y-\sqrt{3} a^2=0

) is correct because it represents the locus of point A such that the difference of angles C and B is π/3.

More Questions from 3D GEOMETRY SYSTEM

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3