The line that goes through the point (0, 3/2 ) and is orthogonal to a tangent line to the part of parabola y =x^2 with x > 0 is y = ax + 3/2.

Question

1. Find the slope $a$ which is in $y=ax+3/2$
2. Find the $x$-coordinate of the intersection of the two lines.

Answer 1

Step by step approach (as requested in comment to question):

Call the $x$-coordinate of the intersection of the tangent and normal lines $x_0$.

The $y$-coordinate is then $y_0=x_0^2$.

Call $m$ the slope of the tangent line to $y=x^2$ at $x_0$. What is $m$ in terms of $x_0$?

What is the slope of the normal line at $x_0$? [Hint: it is perpendicular to the tangent line.]

What is the $y$-intercept of the normal line at $x_0$? [Hint: you just found its slope, and $(x_0,x_0^2)$ is a point on it.]

Given that the $y$-intercept of the normal line is $\dfrac32$ and $x_0>0$, what is $x_0$?

Above you got the slope of the normal line at $x_0$ in terms of $x_0$; this is $a$.

Here is how I picture it, with the parabola in red, the tangent line in blue, and the normal line in green:

Answer 2

Hint: Solve the equation $$x^2-ax-\frac{3}{2}=0$$ and set the discriminant equal to zero and determine the variable $a$.