If P is a point on the parabola y=x^{2}+4 whi | vedclass

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(D) Let the point $P$ on the parabola $y=x^{2}+4$ be $(h, k)$. Since $P$ lies on the parabola,we have $k = h^{2}+4$.The given straight line is $L: y =

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$(2,8)$

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(D) Let the point $P$ on the parabola $y=x^{2}+4$ be $(h, k)$. Since $P$ lies on the parabola,we have $k = h^{2}+4$.The given straight line is $L: y = 4x - 1$,which can be written as $4x - y - 1 = 0$.The perpendicular distance $d$ from point $P(h, k)$ to the line $4x - y - 1 = 0$ is given by:$d = \frac{|4h - k - 1|}{\sqrt{4^{2} + (-1)^{2}}} = \frac{|4h - (h^{2} + 4) - 1|}{\sqrt{16 + 1}} = \frac{|4h - h^{2} - 5|}{\sqrt{17}} = \frac{|-(h^{2} - 4h + 5)|}{\sqrt{17}} = \frac{h^{2} - 4h + 5}{\sqrt{17}}$.To find the point closest to the line,we minimize $d$ by differentiating with respect to $h$:$\frac{dd}{dh} = \frac{1}{\sqrt{17}} (2h - 4)$.Setting $\frac{dd}{dh} = 0$,we get $2h - 4 = 0$,which implies $h = 2$.For $h = 2$,the $y$-coordinate is $k = (2)^{2} + 4 = 4 + 4 = 8$.Thus,the coordinates of point $P$ are $(2, 8)$.

If $P$ is a point on the parabola $y=x^{2}+4$ which is closest to the straight line $y =4 x -1,$ then the coordinates of $P$ are:

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