Parabola
Inscribed Triangle in Parabola
Grade 11
Question:
<p>The area (in sq. units) of an equilateral triangle inscribed in the parabola \(y^2 = 8x\), with one of its vertices on the vertex of this parabola, is</p>
<p>(a) \(64\sqrt{3}\)</p>
<p>(b) \(256\sqrt{3}\)</p>
<p>(c) \(192\sqrt{3}\)</p>
<p>(d) \(128\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Use the symmetry of the parabola about the x-axis to place the equilateral triangle with one vertex at the origin and two vertices symmetric about the x-axis.
<p><strong>Step 1:</strong> Equation of given parabola is \(y^2 = 8x\) and the equilateral triangle inscribed in the given parabola has one vertex on the vertex of the parabola.</p><p><strong>Step 2:</strong> By symmetry, the other two vertices of the equilateral triangle can be taken as \(A(2t^2, 4t)\) and \(B(2t^2, -4t)\).</p><p><strong>Step 3:</strong> Area of \(\triangle OAB = \frac{1}{2}(2t^2)(8t) = 8t^3\).</p><p>∴ Answer is (c) \(192\sqrt{3}\).</p>
Correct Answer: C