Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 79

Show that there are tangent lines to the parabola $y=x^2$ that pass through the point $(0,-4)$. Find
the coordinates of the points where these tangent lines intersect the parabola. Draw the diagram.

By using the equation of the line $y = mx +b$, we know that the given point contains $y$-intercept $b$ as -4.
So the equation is...


$
\begin{equation}
\begin{aligned}
y &= mx +b\\
y &= mx - 4 && \text{Equation 1}
\end{aligned}
\end{equation}
$


Recall that the derivative of the curve is equal to the slope so...


$
\begin{equation}
\begin{aligned}
y &= x^2\\
y ' &= m = 2x
\end{aligned}
\end{equation}
$


Since the lines and the parabola intersects, we can equate its $y$-value and solve for $x$ simultaneously
and find the coordinates of the points we have from Equation 1.


$
\begin{equation}
\begin{aligned}
y & = mx - 4\\
x^2 & = 2x (x) - 4\\
x^2 &= 4\\
x &= \pm \sqrt{4}\\
x &= \pm 2
\end{aligned}
\end{equation}
$


Therefore, the coordinates are $(2,4)$ and $(-2,4)$

Hence, the equation of the tangent lines are

$
\begin{equation}
\begin{aligned}
y-y_1 &= m(x-x_1)
& \phantom{x}&&
y - y_2 &= m(x-x_2)\\
y-y_1 &= 2x_1(x-x_1)
& \phantom{x}&&
y-y_2 &= 2x_2(x-x_2)\\
y - 4 &= 2(2)(x-2)
& \text{and} &&
y-4 &= 2(-2)(x-(-2))\\
y &= 4x - 8 + 4
& \phantom{x}&&
y &= -4x - 8 +4\\
y &= 4x-4
&\phantom{x}&&
y &= -4x-4
\end{aligned}
\end{equation}
$