bobhans-How-do-you-find-a-point-on-the-curve-y-x-2-closest-to-the-point-0-18-

Question Number 106779 by bobhans last updated on 07/Aug/20

^{≻ bobhans ≺}

How do you find a point on the curve y = x^{2}

closest to the point (0, 18) ?

Answered by john santu last updated on 07/Aug/20

^{@ JS @}

let (\sqrt{x}, x) a point on the curve

y = x^{2} and closest to point (0, 18)

so we have d = \sqrt{{(\sqrt{x} - 0)}^{2} + {(x - 18)}^{2}}

d = \sqrt{x + x^{2} - 36 x + 324}

d = \sqrt{x^{2} - 35 x + 324}

put d^{'} = 0 \Rightarrow d^{'} = \frac{2 x - 35}{2 \sqrt{x^{2} - 35 x + 324}} = 0

\to so the coordinates point we

search is (\pm \sqrt{\frac{35}{2}}, \frac{35}{2}) .

Answered by bobhans last updated on 07/Aug/20

^{≻ bobhans ≺}

find tangent to y = x^{2} at (x_{o}, y_{o}) is y = {2 x}_{o} (x - x_{o}) + y_{o}

find normal to y = x^{2} at (x_{o}, y_{o}) is

y = - \frac{1}{{2 x}_{o}} (x - x_{o}) + y_{o} passes through (0, 18)

18 = - \frac{1}{{2 x}_{o}} (- x_{o}) + y_{o} \Rightarrow y_{o} = 18 - \frac{1}{2} = \frac{35}{2}

so we get x_{o} = \pm \sqrt{y_{o}} = \pm \sqrt{\frac{35}{2}}

therefore the point are (\pm \sqrt{\frac{35}{2}}, \frac{35}{2})

Answered by 1549442205PVT last updated on 07/Aug/20

Let A (x, x^{2}) be any point lying on

the curve y = x^{2} . Then

MA = \sqrt{{(x - 0)}^{2} + {(x^{2} - 18)}^{2}}

= \sqrt{x^{4} - {35 x}^{2} + 324} = \sqrt{{(x^{2} - \frac{35}{2})}^{2} + \frac{71}{4}}

⩾ \sqrt{\frac{71}{4}} = \frac{\sqrt{71}}{2} .

The equality ocurrs when x^{2} = \frac{35}{2}

\Leftrightarrow x = \pm \sqrt{\frac{35}{2}} = \pm \frac{\sqrt{70}}{2} \Rightarrow y = x^{2} = \frac{35}{2}

Thus, the point lying on the curve

y = x^{2} which is nearest point to the

point (0; 18) is the point A (\pm \frac{\sqrt{70}}{2}; \frac{35}{2})

and that shortest distance equal to

\frac{\sqrt{71}}{2}

Commented by bemath last updated on 07/Aug/20

typo it should be A (\pm \frac{\sqrt{70}}{2}, \frac{35}{2})

Commented by 1549442205PVT last updated on 07/Aug/20

Thank you sir . I corrected