Let P(4, -4) and Q(9, 6) be two points on the | vedclass

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(B) The area of $\Delta PXQ$ is maximum when the tangent at $X$ is parallel to the chord $PQ$.The slope of the chord $PQ$ is $m = \frac{6 - (-4)}{9 -

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$\frac{125}{4}$

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(B) The area of $\Delta PXQ$ is maximum when the tangent at $X$ is parallel to the chord $PQ$.The slope of the chord $PQ$ is $m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$.The equation of the parabola is $y^2 = 4x$,so $2y \frac{dy}{dx} = 4$,which gives $\frac{dy}{dx} = \frac{2}{y}$.Setting the slope of the tangent equal to the slope of the chord: $\frac{2}{y} = 2 \Rightarrow y = 1$.Since $X$ lies on $y^2 = 4x$,for $y = 1$,we have $1^2 = 4x \Rightarrow x = \frac{1}{4}$. Thus,$X = (\frac{1}{4}, 1)$.The area of $\Delta PXQ$ with vertices $P(4, -4)$,$Q(9, 6)$,and $X(\frac{1}{4}, 1)$ is given by the determinant formula:Area $= \frac{1}{2} |x_P(y_X - y_Q) + x_X(y_Q - y_P) + x_Q(y_P - y_X)|$Area $= \frac{1}{2} |4(1 - 6) + \frac{1}{4}(6 - (-4)) + 9(-4 - 1)|$Area $= \frac{1}{2} |4(-5) + \frac{1}{4}(10) + 9(-5)|$Area $= \frac{1}{2} |-20 + 2.5 - 45| = \frac{1}{2} |-62.5| = 31.25 = \frac{125}{4}$ sq. units.

Let $P(4, -4)$ and $Q(9, 6)$ be two points on the parabola $y^2 = 4x$. Let $X$ be any point on the arc $POQ$ of this parabola,where $O$ is the vertex,such that the area of $\Delta PXQ$ is maximum. Then this maximum area (in sq. units) is

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