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. 

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 $$9/x_0^4+9x_0^2/4=(9/4)(y_0^2+1/y_0).$$
