Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Re: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

We want to minimize the distance $\displaystyle D$ between $\displaystyle (x, x^2+3)$ and $\displaystyle (1,5)$. We know that

$\displaystyle D = \sqrt{(x-1)^2 + (x^2 + 3 - 5)^2} = \sqrt{2x^2 - 6x + 5}$

To make things easier, $\displaystyle D^2 = 2x^2 - 6x + 5 $. To find the x that minimizes $\displaystyle D$, you may as well find the x that minimizes $\displaystyle D^2$. So find the (x,y) coordinate that minimizes $\displaystyle D^2$.

Re: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

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**Roleparadise** Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Let required point be (x,y) Let D= distance from (1,5) So D^2=(x-1)^2+(y-5)^2 But (x,y) is on curve so y=x^2+3 giving D^2=(x-1)^2+(x^2-2)^2

Want D^2 to be minimum (then D will be minimum) So want derivative=0

Hence want 2(x-1)+2(x^2-2)2x=0 2x-2+4x^3-8x=0 4x^3-6x-2=0 Using a numerical change of sign method I get x=1,366 so y=4.866

Re: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

I dont agree with your expression for D

Re: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Whoops, I did $\displaystyle (x-2)^2$ instead of $\displaystyle (x^2-2)^2$...

Anyway, $\displaystyle D^2 = (x-1)^2 + (x^2 - 2)^2 = x^4 - 3x^2 - 2x + 5$. Take derivative with respect to x and set to zero:

$\displaystyle 4x^3 - 6x - 2 = 0$.

There are three real solutions for x: x = -1, -.366, and 1.366. Check each (x,y) point to see which one is closest.

Re: Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

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Originally Posted by

**Roleparadise** Find the point on the parabola y = x^2 + 3 that is closest to the point (1,5)

Here is an entirely different way to solve it

Find a point $\displaystyle (a,b)$ on the curve where the tangent is perpendicular to the normal through $\displaystyle (1,5)$.

Thus solve $\displaystyle \frac{b-5}{a-1}=\frac{-1}{2a}~\&~b=a^2+3$.