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JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of an equilateral triangle inscribed in the parabola \(\mathrm{y}^{2}=8 \mathrm{x},\) with one of its vertices on the vertex of this parabola, is
- A \(64 \sqrt{3}\)
- B \(256 \sqrt{3}\)
- C \(192 \sqrt{3}\)
- D \(128 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(192 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\tan 30^{\circ}=\frac{4 \mathrm{t}}{2 \mathrm{t}^{2}}=\frac{2}{\mathrm{t}} \Rightarrow \mathrm{t}=2 \sqrt{3}\) \(\mathrm{AB}=8 \mathrm{t}=16 \sqrt{3}\) Area \(=256.3 \cdot \frac{\sqrt{3}}{4}=192 \sqrt{3}\)
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