A tangent line to $y=\frac1{x^2}$ cuts the $x$-axis at $A$ and $y$ at $B$, minimize $AB$. The problem: 
A tangent line to $y= \frac{1}{x^2}$ intersects the x-axis at the point A and the y-axis at the point B. 
What is the length of the shortest such line segment AB?
I know that the graph of $y= \frac{1}{x^2}$ looks like a bell centered around the y-axis.
The function is not equal to zero for x at any point, which is why a tangent line (unknown) will cross the x-axis at one point (A), and the y-axis at another point (B). 
The first derivative of $y=\frac{1}{x^2}$ is $\frac{-2}{x^3}$.
I don't know where to go from here. Do I use the first derivative of the function and plug in values A and B as unknown values? I only need guidance on how to begin, and I can solve the rest from there. Thank you. 
 A: Hint:  For each $x$, you know the line through the function goes through the point $(x, y) = \left(x, \frac{1}{x^2}\right)$ and has slope $\frac{-2}{x^3}$.  Can you find the $x$-intercept and the $y$-intercept of such a line and then use the distance formula to find its length?  (There could be a convenient way to write the formula of a line when you know a point and the slope.)
You still have the minimization to do, but first you must be able to turn each point on the curve into the length of the line through it.
A: Some preliminary theory: If a line tangent to a given function intersects the x-axis at a point A and the y-axis at a point B, then it must form a right triangle with the right angle situated at the origin and the two sides with vertices located at $(A, 0)$ and $(0, B)$ of lengths $A$ and $B$ respectively. The length of the hypotenuse of a right triangle thus formed is the line segment $\overline{AB}$.
First of all, let's find the equation of the line tangent to the function $y=\frac{1}{x^2}$ at an arbitrary point $\alpha$ in general:
$$
y - f(\alpha)=f'(\alpha)(x-\alpha)\implies\\
y - \frac{1}{\alpha^2}=-\frac{2}{\alpha^3}(x - \alpha)\implies\\
y=-\frac{2}{\alpha^3}x+\frac{3}{\alpha^2}
$$
Now, in the problem it is stated that when $x=A$, $y=0$ and when $x=0$, $y=B$. So, plugging in these two pairs of values for $x$ and $y$, we will work out the lengths of the two legs of the right triangle expressed in terms of our arbitrary point $\alpha$.
The length of the side along the x-axis:
$$
0=-\frac{2}{\alpha^3}A+\frac{3}{\alpha^2}\implies\\
A=\frac{3\alpha}{2}
$$
The length of the side along the y-axis:
$$
B=-\frac{2}{\alpha^3}\cdot 0+\frac{3}{\alpha^2}\implies\\
B=\frac{3}{\alpha^2}
$$
Now we can construct a function $L$ which will be the length of the hypotenuse of our imaginary right triangle expressed in terms of our point $\alpha$:
$$
L(\alpha)=\sqrt{A^2+B^2}=\sqrt{\left(\frac{3\alpha}{2}\right)^2+\left(\frac{3}{\alpha^2}\right)^2}=\frac{3\sqrt{\alpha^8+4}}{2\alpha^2}
$$
So, here's what we've got. That function $L$ is a function that can give us any possible value for the length of the line segment $\overline{AB}$ including the one that's the shortest. All we need to do is to plug in the right number for $\alpha$. So, we need to find an $\alpha$ such that when we plug it into the function, it will spit out the functional value that's the smallest of all possible values that the function can yield. In order to do that, we need to find its first derivative, set it equal to zero and solve for $\alpha$. That alpha is that magic number which will give us the shortest hypotenuse.
$$
L'(\alpha)=\left(\frac{3\sqrt{\alpha^8+4}}{2\alpha^2}\right)'=\frac{6\alpha^5}{\sqrt{\alpha^8+4}}-\frac{3\sqrt{\alpha^8+4}}{\alpha^3}
$$
$$
\frac{6\alpha^5}{\sqrt{\alpha^8+4}}-\frac{3\sqrt{\alpha^8+4}}{\alpha^3}=0\\
\alpha=\pm\sqrt[4]{2}
$$
This means that there are two such points. Now, let's plug these points back into the length function and find out the shortest hypotenuses:
$$
L(\sqrt[4]{2})=\frac{3\sqrt{(\sqrt[4]{2})^8+4}}{2(\sqrt[4]{2})^2}=3\\
L(-\sqrt[4]{2})=\frac{3\sqrt{(-\sqrt[4]{2})^8+4}}{2(-\sqrt[4]{2})^2}=3
$$
The lengths as expected are positive values and actually the same.
Answer: the length of the shortest such line segment $\overline{AB}$ is $3$ and there are actually two such line segments.
A: The derivative of $y=1/x^2$ with respect to $x$ is $-2/x^3$ so the equation of the  tangent-line at $(x_0,y_0)$ is $$(y-y_0)=(-2/x_0^3)(x-x_0).$$ In this equation, when $x=0$ we have $(y-1/x_0^2)=(y-y_0)=(-2/x_0^3)(0-x_0)=2/x_0^2 .$ Hence,  $y=3/x_0^2.$ So the tangent-line at $(x_0,y_0)$ meets the $y$-axis at $$P=(0,3/x_0^2).$$ In the equation for the tangent-line, when $y=0$ we have $-1/x_0^2=(0-1/x_0^2)=(y-1/x_0^2)=(-2/x_0^3)(x-x_0)$, which implies $x=3x_0/2.$ So the tangent-line meets the $x$-axis at $$Q=(3x_0/2,0).$$ Now find the value(s) of $x_0$ (or of $y_0)$ that minimize the square of the distance from $P$ to $Q.$ The square of the distance from $P$ to $Q$ is $$9/x_0^4+9x_0^2/4=(9/4)(y_0^2+1/y_0).$$
