Finding the equation of the line that is orthogonal to a tangent line to a Parabola y = x² There is a parabola $y = x^2$. The question ask for the "a" of the equation $y = ax +$$3 \over 2$, which is the equation for the line that goes through the point $(0,$$3 \over 2$$)$ and is orthogonal to a tangent line of the parabla above, with $x > 0$.
The only problem for me is to understand of which tangent line, from infinite (as far as I know) when x > 0 is he talking about. The asnwer for this is $-$$1 \over 2$ which implies the tangent line being equal to $y = 2x -1$.
As far as I know, I could also choose the tangent line $y = x -$$1 \over 4$, then the orthogonal line would be $y=-2.5x+$$3\over2$, and it would still match the requirements of the question.

The generic tangent line to the parabola reads
$$
y = y_0 +2x_0(x-x_0)
$$
where $y_0 = x_0^2$
the the normal line should read 
$$
y = \frac{3}{2}-\frac{1}{2x_0}x
$$
Their intersection point is the solution for
$$
\left\{
\begin{array}{rcl}
y & = & y_0 +2x_0(x-x_0)\\
y & = & \frac{3}{2}-\frac{1}{2x_0}x
\end{array}
\right.
$$
giving
$$
x_0 = \frac{x_0 (3 + 2 x_0^2)}{1 + 4 x_0^2}\\
y_0 = \frac{5x_0^2}{1+4x_0^2}
$$
but $$y_0 = x_0^2 \Rightarrow \frac{4 x_0^2 \left(x_0^2-1\right)^2}{\left(4 x_0^2+1\right)^2} = 0 \Rightarrow x_0 = \pm 1
$$

