Find-the-point-on-the-curve-x-2y-2-closest-to-the-point-10-0-

Question Number 121264 by benjo_mathlover last updated on 06/Nov/20

Find the point on the curve

x = {2 y}^{2} closest to the point

(10, 0)

Answered by liberty last updated on 06/Nov/20

Let L be the square of distance between

(x, y) and (10, 0) .

L = {(x - 10)}^{2} + y^{2} = {({2 y}^{2} - 10)}^{2} + y^{2}

L = {4 y}^{4} - {39 y}^{2} + 100

\frac{dL}{dy} = {16 y}^{3} - 78 y = 0

2 y ({8 y}^{2} - 39) = 0, y = 0; \pm \frac{\sqrt{39}}{2 \sqrt{2}}

\frac{dL}{dy} < 0 on (- \infty, - \frac{\sqrt{39}}{2 \sqrt{2}}) and (0, \frac{\sqrt{39}}{2 \sqrt{2}})

\frac{dL}{dy} > 0 on (- \frac{\sqrt{39}}{2 \sqrt{2}}, 0) and (\frac{\sqrt{39}}{2 \sqrt{2}}, \infty)

When y = - \frac{\sqrt{39}}{2 \sqrt{2}}, x = \frac{39}{4}

and y = \frac{\sqrt{39}}{2 \sqrt{2}}, x = \frac{39}{4} . ∵ The points are

(\frac{39}{4}, - \frac{\sqrt{39}}{2 \sqrt{2}}) and (\frac{39}{4}, \frac{\sqrt{39}}{2 \sqrt{2}})

Answered by mr W last updated on 06/Nov/20

s a y p o i n t P (2 p^{2}, p)

1 = 4 y \frac{d y}{d x}

\frac{d y}{d x} = \frac{1}{4 p} = - \frac{10 - 2 p^{2}}{0 - p}

\Rightarrow - 8 p^{2} + 40 = 1

\Rightarrow p = \pm \frac{\sqrt{39}}{2 \sqrt{2}}

\Rightarrow P (\frac{39}{4}, \pm \frac{\sqrt{39}}{2 \sqrt{2}})

Commented by liberty last updated on 06/Nov/20

it should be 1 = 4 y \frac{dy}{dx}

Commented by mr W last updated on 06/Nov/20

y e s . t h a n k s!

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