Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 100

Show two perpendicular lines that intersect on the $y$-axis and are both tangent to the parabola
$y = x^2$ by illustrating its diagram. State where do these lines intersect.




Since $y=x^2$ is an even function and the lines intersect on the $y$-axis and are perpendicular to each other;
the line makes an angle of $45^\circ$ to the positive and negative $x$-axis. Recall that the tangent of the
angle is equal to the slope of the line so,

$m = \tan(\pm 45^\circ)$
$m = \pm 1$

Also, the derivatives of the curve is equal to the slope so

$\displaystyle \frac{dy}{dx} = m \frac{dy}{dx}(x^2)$


$\begin{equation} \begin{aligned} m &= 2x\\ m &= \pm 1\\ x &= \pm \frac{1}{2}\\ \end{aligned} \end{equation}$


Solving for $y$,


$\begin{equation} \begin{aligned} y &= \left( \pm \frac{1}{2}\right)^2\\ y &= \frac{1}{4} \end{aligned} \end{equation}$


Now, using point slope form to get the equation of the line
$y-y_1 = m(x-x_1)$


$\begin{equation} \begin{aligned} y - \frac{1}{4} &= 1 \left( x- \frac{1}{2} \right) && \text{and} && y - \frac{1}{4} = -1 \left( x - \left( - \frac{1}{2} \right) \right) \\ y - \frac{1}{4} &= x - \frac{1}{2} && \phantom{x} && y - \frac{1}{4} = -x - \frac{1}{2} \\ y &= x - \frac{1}{2} + \frac{1}{4} && \phantom{x} && y = -x - \frac{1}{2} + \frac{1}{4}\\ y &= x - \frac{1}{4} && \phantom{x} && y = -x-\frac{1}{4}\\ \end{aligned} \end{equation}$


We can get the intersection by using the equations of the line we obtain

$\begin{equation} \begin{aligned} x - \cancel{\frac{1}{4}} &= -x - \cancel{\frac{1}{4}}\\ 2x &= 0\\ x &= 0 \end{aligned} \end{equation}$

Solving for $y$,


$\begin{equation} \begin{aligned} y &= 0 - \frac{1}{4}\\ y &= \frac{1}{4} \end{aligned} \end{equation}$


Therefore, the line intersects at point $\displaystyle\left(0,\frac{1}{4}\right)$ as shown from the graph