Question: Find the normal equation to the graph $ y = 2x^2$, that goes through the point $ (1,0.25)$. How this is done? Im not even really sure what I'm being asked here...

Answer: 
The Question
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is to find a line passing through the point $(1,0.25)$ (doesn't lie on the graph) and perpendicular to the tangent line to the $y=2x^2$ at the point of intersection.

Solution
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The required line passing through the point $(1,0.25)$ and a point $(a,2a^2)$ of the graph. The slope of the tangent line at a point $(a,b)=(a,2a^2)$ is $y'(a)=4a.$
Now the slope of the normal is
$$\frac{-1}{4a}=\frac{2a^2-0.25}{a-1}$$
Solving for $a$, we get $a=0.5$ and find the slope is $-0.5$ and the line equation is
$$y-0.25=-0.5(x-1)$$
Draw a picture of the curve $y=2x^2$. Locate the point $(1.0.25)$. Notice that the point is *not* on the curve. 

You want to draw a line through $(1,0.25)$ which meets the parabola at right angles. 

We outline a path that will lead you to the answer.

1) Let the point at which the line we are looking for meets the parabola be $(a,2a^2)$. 

2) Find an expression for the slope of the tangent line at $x=a$ in terms of $a$.

3) Use 2) to find an expression for the slope of a line perpendicular to the tangent line.

4) You now the slope of the normal, and you know it passes through $(a,2a^2)$. So you can find the equation of the normal, *in terms of* the unknown $a$.

5) Use the fact that this line passes through $(1,0.25)$ to write down an equation for $a$, and solve. 

6) Now that you know $a$, you know the equation of the line we are looking for.