Question:
Let \(P(4, -4)\) and \(Q(9, 6)\) be two points on the parabola, \(y^2 = 4x\) and let \(x\) be any point on the arc \(POQ\) of this parabola, where \(O\) is the vertex of this parabola, such that the area of \(\Delta PXQ\) is maximum. Then this maximum area (in sq. units) is :
Let \(P(4, -4)\) and \(Q(9, 6)\) be two points on the parabola, \(y^2 = 4x\) and let \(x\) be any point on the arc \(POQ\) of this parabola, where \(O\) is the vertex of this parabola, such that the area of \(\Delta PXQ\) is maximum. Then this maximum area (in sq. units) is :
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The area of \(\Delta PXQ\) is maximum when the tangent at \(X\) is parallel to chord \(PQ\) (slope 2). Find \(X\) and compute the area.
Updated On: Nov 5, 2026
- \(\frac{625}{4}\)
- \(\frac{125}{4}\)
- \(\frac{75}{2}\)
- \(\frac{125}{2}\)
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The Correct Option is B
Solution and Explanation
Slope of \(PQ = \frac{6-(-4)}{9-4} = 2\). The area of \(\Delta PXQ\) is largest when the tangent at \(X\) on arc \(POQ\) is parallel to \(PQ\).
For \(y^2 = 4x\), take \(X = (t^2, 2t)\); tangent slope \(= \frac{dy}{dx} = \frac{1}{t} = 2 \Rightarrow t = \frac12\), so \(X = \left(\frac14, 1\right)\). (\(P\) has \(t=-2\), \(Q\) has \(t=3\), so \(t=\frac12\) lies on arc \(POQ\).)
Area \(= \frac12\left|4(6-1) + 9(1-(-4)) + \frac14(-4-6)\right| = \frac12\left|20 + 45 - \frac{5}{2}\right| = \frac12\cdot\frac{125}{2} = \frac{125}{4}\).
Answer: (2) \(\frac{125}{4}\)
For \(y^2 = 4x\), take \(X = (t^2, 2t)\); tangent slope \(= \frac{dy}{dx} = \frac{1}{t} = 2 \Rightarrow t = \frac12\), so \(X = \left(\frac14, 1\right)\). (\(P\) has \(t=-2\), \(Q\) has \(t=3\), so \(t=\frac12\) lies on arc \(POQ\).)
Area \(= \frac12\left|4(6-1) + 9(1-(-4)) + \frac14(-4-6)\right| = \frac12\left|20 + 45 - \frac{5}{2}\right| = \frac12\cdot\frac{125}{2} = \frac{125}{4}\).
Answer: (2) \(\frac{125}{4}\)
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