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The point on the parabola $y^{2}=64 x$ which is nearest to the line $4 x+3 y+35=0$ has coordinates
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Verified Answer
The correct answer is:
(9,-24)
Given equation of parabola is
$$
y^{2}=64 x
$$
The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.
On differentiating both sides of Eq. (i), we get $2 y \frac{d y}{d x}=64$
$\Rightarrow \quad \frac{d y}{d x}=\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$.
$\therefore$
$$
-\frac{4}{3}=\frac{32}{y} \Rightarrow y=-24
$$
From Eq. (i), $(-24)^{2}=64 x \Rightarrow x=9$
$\therefore$ Hence, the required point is (9,-24)
$$
y^{2}=64 x
$$
The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.
On differentiating both sides of Eq. (i), we get $2 y \frac{d y}{d x}=64$
$\Rightarrow \quad \frac{d y}{d x}=\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$.
$\therefore$
$$
-\frac{4}{3}=\frac{32}{y} \Rightarrow y=-24
$$
From Eq. (i), $(-24)^{2}=64 x \Rightarrow x=9$
$\therefore$ Hence, the required point is (9,-24)
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