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.
 A: 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 $$y_0 = x_0^2 \Rightarrow \frac{4 x_0^2 \left(x_0^2-1\right)^2}{\left(4 x_0^2+1\right)^2} = 0 \Rightarrow x_0 = \pm 1
$$

A: Hint: Let $a$ = slope of the line you are finding, then the slope of the tangent line will be $- \frac{1}{a}$.
$f(x) = x^2$ then the derivative would be $f'(x) = 2x$.
In the equation of the generic tangent line of a parabola, use the part of that that represent the slope, $f'(x)$ as the slope of the tangent line.
We can say that point $(x_1, y_1)$ will be the intersection of this two line and the parabola.
To get then slope of the tangent line you will need to substitute the x-coordinate of $(x_1, y_1)$ to the $f'(x)$. Since $- \frac{1}{a}$ is the slope of the tangent line also, then $f'(x1) = - \frac{1}{a}.$
Substituting $f'(x_1)$ by $2(x_1)$, then $2(x_1)= -\frac{1}{a}$. By equating we get, $x_1 = - \frac{1}{2a}$.
Substitute this $x_1$ by the $y = ax + \frac{3}{2}$, we will get $y_1 = 1$. Since this point intersect the parabola $y = x^2$, substitute $y$ by the equation and equate it, you will get $x = \pm 1$. But we are talking about the parabola on $x > 0$. Get the positive value.
$(1,1)$ is the point we get from equating all of this. Substitute this point in  $y = ax + \frac{3}{2}$ we will get $a= - \frac{1}{2}$.
A: HINT


*

*find the intersection point between the line and the parabola for $x>0$ as a function of the unknown coefficient $a$, that is


$$x^2=ax+\frac32\iff x^2-ax+\frac32=0 \iff x=\frac{a \pm\sqrt{a^2-6}}{2}\implies x_0=\frac{a +\sqrt{a^2-6}}{2}>0$$


*

*set the orthogonality condition at the intersection point to find $a$ value


$$y=x^2\implies y'(x_0)=2x_0=-\frac1a\implies a +\sqrt{a^2-6}=-\frac1a$$
A: First of all you put $x=c$ (constant) 
then find out the general equation of normal line.
it will be 
$$x+2cy = c +2c^3$$
Now this line passes through $(0,1.5)$
Find out the value of $c$ .
Then find out the slope of normal which is $\frac{1}{2c}=a$
