OAB is an equilateral triangle where O- 0-0 -A- 1- 1 -3- The co-ordinates of point B can be -

If AB is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ such that $Δ O A B$ is an equilateral triangle O being the origin, then the eccentricity of the hyperbola satisfies

Q. The co-ordinates of vertices

P

and

Q

of an equilateral

Δ

are

(1, \sqrt{3})

and

(0, 0)

. Which of the following could be co-ordinates of

R

Q. Let

P (sin θ, cos θ)

, where

0 \leq θ \leq 2 π

be a point and let

O A B

be a triangle with vertices

(0, 0), (\frac{\sqrt{3}}{2}, 0)

and

(0, \frac{\sqrt{3}}{2})

. Find

θ

P

lies inside the

△ O A B

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△OAB is an equilateral triangle where O≡(0,0),A≡(1,1√3). The co-ordinates of point B can be :-

The correct option is C (1,−1√3)Lenth of sides is  ⎷(1−0)2+(1√3−0)2=√1+13=√43Then from options only point (1,−1√3) gives the same length

$△ O A B$ is an equilateral triangle where $O \equiv (0, 0), A \equiv (1, \frac{1}{\sqrt{3}})$ . The co-ordinates of point $B$ can be :-

The correct option is C $(1, - \frac{1}{\sqrt{3}})$
Lenth of sides is
$\sqrt{{(1 - 0)}^{2} + {(\frac{1}{\sqrt{3}} - 0)}^{2}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}}$
Then from options only point $(1, - \frac{1}{\sqrt{3}})$
gives the same length