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

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