Determine the points on the parabola $y=x^2 - 25$ that are closest to $(0,3)$ 
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.
 A: 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.
A: 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 $r$.
A: 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 $x^2-\frac{55}2=0\implies x=\pm\sqrt{\frac{55}2}, y=x^2-25=\frac{55}2-25=\frac52$
A: 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.
