TS EAMCET · Maths · Parabola
If one of the vertices of an equilateral triangle inscribed in the parabola \(y^2=12 x\) coincides with the vertex of the parabola, then the area (in sq. units) of that triangle is
- A \(192 \sqrt{3}\)
- B \(864 \sqrt{3}\)
- C \(216 \sqrt{3}\)
- D \(432 \sqrt{3}\)
Answer & Solution
Correct Answer
(D) \(432 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(y^2=12 x\) \(\begin{aligned} & \Rightarrow \quad a^2 \sin ^2 30^{\circ}=12 \cdot a \cos 30^{\circ} \\ & \Rightarrow \quad a=48 \cos 30^{\circ}=24 \sqrt{3} \\ & \text { Area }=\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4}(24 \sqrt{3})^2=432 \sqrt{3} .\end{aligned}\)
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