# Find the points on the parabola $\,y=x^2$ that are closest to the point $\,(0,8)$

Find the points on the parabola $y=x^2$ that are closest to the point $(0,8)$.

Having trouble figuring out where to start.

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It's a shame that everybody is showing off ways to actually do the problem, rather than give advice on how to start problems. –  Hurkyl Jul 17 '14 at 2:37

Assume the point is $(x,y)$ then find the distance as

$$d=\sqrt{(x-0)^2+(y-8)^2}= d=\sqrt{x^2+(x^2-8)^2}.$$

Now you can use derivative techniques to find the minimum of the above function.

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I'd recommend minimizing the square of the distance instead. It yields the same result without having to deal with yucky square roots. –  Kaj Hansen Jul 17 '14 at 1:44

Take the parametric point on the parabola $(t,t^2)$, and find the $t$ that minimize the distance-squared between $(t,t^2)$ and $(0,8)$. That is, you need to minimize $d^2=t^2+(t^2-8)^2$ w.r.t $t$. First order condition gives, $2t+4t(t^2-8)=0$. This gives either $t=0$ or $t^2=7/2$. Check which one has lower distance, should be the 2nd one.

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# Method of normal-line

Given is a point $(x,x^2)$. The normal line is given by $(\Delta x, \Delta y) = (-x,\tfrac{1}{2})$. And the normal line will cut the point $(0,8)$, so we obtain

$$x^2 + \frac{1}{2} = 8 \Rightarrow x = \pm \sqrt{\frac{15}{2}}.$$

So the points are given by

$$\left( \pm \sqrt{\frac{15}{2}}, \frac{15}{2}\right).$$.

# Method of minimizing the distance

The distance between a point $(x,x^2)$ on the parabola and the point $(0,8)$ is given by

$$L(x) = \sqrt{ \Big( x - 0 \Big)^2 + \Big( x^2 - 8 \Big)^2 }.$$

The minimum distance is obtained for

$$\frac{dL}{dx} = 0 \wedge \frac{d^2L}{dx^2} > 0.$$

$$\frac{dL}{dx} = \frac{ 2 x^3 - 15 x } {\sqrt{ x^2 + \Big( x^2 - 8 \Big)^2 }}.$$

Thus

$$\frac{dL}{dx} = 0 \Rightarrow x = 0 \vee x = \pm \sqrt{15/2}.$$

$$\frac{d^2L}{dx^2} = \frac{ 6 x^6 - 45 x^4 + 384 x^2 - 960 } {\sqrt{ x^2 + \Big( x^2 - 8 \Big)^2 }},$$

whence

$$\begin{eqnarray} x = 0 & \Rightarrow & \frac{d^2L}{dx^2} < 0 \Rightarrow \textrm{local maximum}\\ x = \pm \sqrt{\frac{15}{2}} & \Rightarrow & \frac{d^2L}{dx^2} > 0 \Rightarrow \textrm{minimum}. \end{eqnarray}$$

So we obtain then the points

$$\begin{eqnarray} \left( \pm \sqrt{\frac{15}{2}}, \frac{15}{2}\right) & \Rightarrow & L = \frac{1}{2} \sqrt{ 31 }. \end{eqnarray}$$

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One way of doing this problem without calculus is to place the center of a circle at (0,8), and gradually expand the circle. Eventually, the circle will touch the parabola at exactly 2 points (because the circle is centered on the parabola's axis of symmetry). These points are (-c,d), and (c,d). Notice that vertical component (aka $y$) only has a single solution (d).

To find the solution, we have to solve the following system. $$x^2+(y-8)^2=r^2$$ $$y=x^2$$ substiuting we have $$y + (y-8)^2=r^2$$ $$y+(y^2-16y+64)-r^2=0$$ $$y^2-15y+(64-r^2)=0$$ $$y=\frac{15\pm\sqrt{15^2-4(64-r^2)}}{2}$$ Recall that we said that $y$ has a single solution, which means that the discriminant ($15^2-4(64-r^2)$) is zero. Thus we have $$y=\frac{15\pm\sqrt{0}}{2}=\frac{15}{2}$$ and $$x=\pm\sqrt{\frac{15}{2}}$$ so the points are $(-\sqrt{15/2},15/2)$ and $(\sqrt{15/2},15/2)$

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Suppose that the closest point is at $(p,p^2)$. Then the line connecting this point to $(0,8)$ must be normal to the curve. Since $\frac{d}{dx} x^2 = 2x$, we know that the slope of the tangent line at $(p,p^2)$ is $2p$, so the slope of the normal line must be the negative reciprocal: $\frac{-1}{2p}$ (here we're assuming that $p \neq 0$). Thus, the equation of the normal line using point-slope form is: $$y - p^2 = \frac{-1}{2p}(x - p)$$ But this line goes through the point $(0,8)$, so we may substitute to solve for $p$: \begin{align*} 8 - p^2 &= \frac{-1}{2p}(0 - p) \\ 8 - p^2 &= \frac{1}{2} \\ 16 - 2p^2 &= 1 \\ 15 &= 2p^2 \\ \frac{15}{2} &= p^2 \\ p &= \pm\sqrt{\frac{15}{2}} \end{align*}

Indeed, both $(-\sqrt{\frac{15}{2}}, \frac{15}{2})$ and $(\sqrt{\frac{15}{2}}, \frac{15}{2})$ are closer to $(0,8)$ than $(0,0)$ (so it was safe to assume that $p \neq 0$).

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Frequently, an excellent starting point for a word problem is to turn it into an algebraic one.

Your searching for points in the plane that have a particular property. So you introduce new variables to represent the coordinates of such a point, and then you try to turn those properties into equations.

Let's call the point that we're solving for $(u,v)$.

You want this point to be on the parabola, which means we have the equation $v = u^2$.

You are asking something about the distance between $(u,v)$ and the point $(0,8)$: that is

$$\sqrt{u^2 + (v-8)^2}$$

Finally, your task is to find the $u$ and $v$ that minimize this formula. So now we have our algebra problem:

Find the values of $(u,v)$ satisfying $v=u^2$ that give the minimum value to the expression $\sqrt{u^2 + (v-8)^2}$,

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With Lagrange Multipliers, we get $$(x,y-8)=\lambda\, (2x,-1) \\ \lambda=1/2 \\ y=8-1/2=15/2$$ and $$x=\pm\sqrt y=\pm \sqrt\frac{15}{2}$$ so the points are $$\left( \pm \sqrt{\frac{15}{2}}, \frac{15}{2}\right).$$

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The distance, $d$, between points $(x,y)$ and $(0,8)$ is given by

$$d = \sqrt{(x-0)^2 + (y-8)^2} = \sqrt{x^2 + ( y - 8 )^2}$$

A common conceptual simplification for these problems is to consider the square of this distance, because a (non-negative) value is minimized when its square is minimized. So, writing $s$ for the squared distance, we get a less-complicated expression $$s = x^2 + ( y - 8 )^2 \qquad (\star)$$

From here, we can introduce the fact that $x$ and $y$ are related to each other: $y=x^2$. Thus, we can reduce the number of variables in $(\star)$. Most answers seem to be suggesting to re-write $y$ in terms of $x$ and apply various Calculus-based minimization techniques (after all, you have tagged your question with calculus); however, it turns out that if you re-write $x$ in terms of $y$ ...

$$s = y + ( y - 8 )^2 \qquad (\star\star)$$

... then minimization is a matter of algebra. (You might note that $(\star\star)$ itself can be viewed as the equation of an upward-opening parabola whose vertex is not difficult to locate.)

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