Finding the equation of the line that is orthogonal to a tangent line to a Parabola y = x²

Question

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.

Answer 1

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

Answer 2

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

$$y=x^2\implies y'(x_0)=2x_0=-\frac1a\implies a +\sqrt{a^2-6}=-\frac1a$$

Answer 3

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 $\frac{1}{2c}=a$

Answer 4

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 $a= - \frac{1}{2}$.