Question ID: #838
An equilateral triangle OAB is inscribed in the parabola $y^2 = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is:
- (1) $4(3-\sqrt{3})$
- (2) $2(8-3\sqrt{3})$
- (3) $4(6+\sqrt{3})$
- (4) $2(3+\sqrt{3})$
Solution:
Let the vertices of the triangle be $O(0,0)$, $A(at^2, 2at)$, and $B(at^2, -2at)$ due to symmetry.
For $y^2 = 4x$, $a = 1$. So $A(t^2, 2t)$.
Since $\Delta OAB$ is equilateral, the angle made by OA with the x-axis is $30^\circ$.
$$ \tan 30^\circ = \frac{y_A}{x_A} = \frac{2t}{t^2} = \frac{2}{t} $$
$$ \frac{1}{\sqrt{3}} = \frac{2}{t} \Rightarrow t = 2\sqrt{3} $$
Coordinates of A:
$$ x = t^2 = (2\sqrt{3})^2 = 12 $$
$$ y = 2t = 4\sqrt{3} $$
Point A is $(12, 4\sqrt{3})$ and B is $(12, -4\sqrt{3})$.
The circle has AB as diameter.
Centre of circle (C) = Midpoint of AB = $(12, 0)$.
Radius of circle (r) = Half of length AB = $y$-coordinate of A = $4\sqrt{3}$.
Distance of the circle from the origin is the distance from the origin to the centre minus the radius (since the origin is outside the circle, $12 > 4\sqrt{3} \approx 6.92$).
$$ d = OC – r = 12 – 4\sqrt{3} $$
$$ d = 4(3 – \sqrt{3}) $$
Ans. (1)
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