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