The point on the parabola y^{2}=64x which is | vedclass

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(A) The given equation of the parabola is $y^{2} = 64x$ $(i)$.The point on the parabola nearest to the line $4x + 3y + 35 = 0$ is the point where the

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$(9, -24)$

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(A) The given equation of the parabola is $y^{2} = 64x$ $(i)$.The point on the parabola nearest to the line $4x + 3y + 35 = 0$ is the point where the tangent is parallel to the given line.The slope of the line $4x + 3y + 35 = 0$ is $m = -\frac{4}{3}$.Differentiating $(i)$ with respect to $x$, we get $2y \frac{dy}{dx} = 64$, which implies $\frac{dy}{dx} = \frac{32}{y}$.Since the tangent is parallel to the line, their slopes must be equal:$\frac{32}{y} = -\frac{4}{3} \Rightarrow y = -24$.Substituting $y = -24$ into $(i)$:$(-24)^{2} = 64x$ $\Rightarrow 576 = 64x$ $\Rightarrow x = 9$.Thus, the required point is $(9, -24)$.

The point on the parabola $y^{2}=64x$ which is nearest to the line $4x+3y+35=0$ has coordinates

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