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Determine the points on the parabola $y=x^2 - 25$ that are closest to $(0,3)$

I would like to know how to go about solving this. I have some idea of solving it. I believe you have to use implicit differentiation and the distance formula but I don't know how to set it up. Hints would be appreciated.

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4 Answers 4

up vote 4 down vote accepted

Just set up a distance squared function:

$$d(x) = (x-0)^2 + (x^2-25-3)^2 = x^2 + (x^2-28)^2$$

Minimize this with respect to $x$. It is easier to work with the square of the distance rather than the distance itself because you avoid the square roots which, in the end, do not matter when taking a derivative and setting it to zero.

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Probably worth noting that $$d(x)=(x-0)^2+(y-3)^2$$ where $y=x^2-25$... –  Thomas Andrews Mar 28 '13 at 14:38
    
So I can just find the derivative of $d(x)$ and I'll be fine? –  Jeel Shah Mar 28 '13 at 14:41
    
@gekkostate: yes. But you'll need to compute the y-coordinates corresponding to the values of $x$ you find from the minimizing. –  Ron Gordon Mar 28 '13 at 14:42
    
@RonGordon Sorry, can you clarify please? –  Jeel Shah Mar 28 '13 at 14:44
    
Find your values of $x$, but the problem is asking for the points, so you should remember to compute the $y$ that goes to each $x$, i.e. $y=x^2-25$. –  Ron Gordon Mar 28 '13 at 14:45

Hint: The formula for a circle around $(0,3)$ is $(y-3)^2 + x^2 = r$

Using this information, you want to find the intersection between $(y-3)^2 + x^2 = r$ and $y=x^2-25$ that minimizes $r$.

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A point $P(x_0,y_0)$ on the parabola, solution of your problem, is such that the point $(0,3)$ is on the normal to the parabola at $P$. The equation of the normal is $\vec{T} \cdot \vec{PM} = 0$, where $M(x,y)$ is a moving point on this line, and $\vec{T}$ is the tangent vector at $P$, thus $\vec{T}=(1,2x_0)$.

The equation can be written: $$1 \cdot (x - x_0) + 2x_0 \cdot (y - y_0) = 0$$ And since $(0,3)$ is on the normal, you must have $$ 1 \cdot (0 - x_0) + 2x_0 \cdot(3-y_0) = 0$$ $$ -x_0+2x_0(3-x_0^2+25) = 0$$ $$ x_0(55-2x_0^2)=0$$ Now it should not be difficult to conclude.

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So, $y=x^2-25$

So, we need to minimize $$f(x)=(x-0)^2+(x^2-25-3)^2=x^4-55x^2+28^2$$

$$\text{So,} f(x)=\left(x^2-\frac{55}2\right)^2+28^2-\left(\frac{55}2\right)^2\ge 28^2-\left(\frac{55}2\right)^2=\frac14$$ as $x$ is real

So, the distance will the minimum if $x^2-\frac{55}2=0\implies x=\pm\sqrt{\frac{55}2}, y=x^2-25=\frac{55}2-25=\frac52$

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@ThomasAndrews, Earlier, I used the common pattern to introduce parametric equation. –  lab bhattacharjee Mar 28 '13 at 14:58
    
From Ron Gordon approach or mine, I find in both cases $x^2=\frac{55}{2}$. Actually, I wonder how factoring a quadratic with $55$ as middle term can yield $27\over2$ ;-) –  Jean-Claude Arbaut Mar 28 '13 at 15:02
    
@arbautjc, if one understands my method, the type error should have been easily discernible which is now rectified. –  lab bhattacharjee Mar 28 '13 at 15:06
    
My instructors always said that :-) –  Jean-Claude Arbaut Mar 28 '13 at 15:07

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