Question 11

Mathematics Coordinate Geometry Hard

An equilateral triangle is inscribed in a parabola \( y^{2}=8 x \) whose one vertix is at the vertex of the parabola then the length of the side of the triangle is:

(A) \( 8 \sqrt{3} \) units
(B) \( 16 \sqrt{3} \) units
(C) \( 4 \sqrt{3} \) units
(D) \( \sqrt{3} / 2 \) units
View Dynamic Solution & Explanation
Correct Solution: Option B

Step-by-step Solution:

The given equation of the parabola is: \[ y^2 = 8x \] The vertex of the parabola is at \( (0,0) \). Since the triangle is equilateral, the axis of the parabola bisects the angle. This gives an angle of \(30^\circ\) above the x-axis. If the side of the equilateral triangle intersects the parabola at \( (2t^2, 4t) \), then we use the tangent condition: \[ \tan 30^\circ = \frac{4t}{2t^2} \] \[ \frac{1}{\sqrt{3}} = \frac{4t}{2t^2} \] Solving for \( t \): \[ t = 2\sqrt{3} \] Now, substituting \( t = 2\sqrt{3} \): \[ x = 2(2\sqrt{3})^2 = 2(12) = 24 \] \[ y = 4(2\sqrt{3}) = 8\sqrt{3} \] Thus, the point of intersection is: \[ (24, 8\sqrt{3}) \] Finding the length of the side: Using the distance formula: \[ \text{Length} = \sqrt{(24 - (-24))^2 + (8\sqrt{3} - (-8\sqrt{3}))^2} \] \[ = \sqrt{(48)^2 + (16\sqrt{3})^2} \] \[ = \sqrt{576 + 192} \] \[ = \sqrt{768} \] \[ = 16\sqrt{3} \] Thus, the side length of the equilateral triangle is: \[ \boxed{16\sqrt{3}} \]