Finding all points on a graph with tangent lines passing through a particular point?

Question

In this particular case, I am trying to find all points $(x,y)$ on the graph of $f(x)=x^2$ with tangent lines passing through the point $(3,8)$.

Now then, I know the graph of $x^2$. What now?

Answer 1

Step 1: For a point $(x,y)$ on the graph of $f(x)=x^2$, find the slope of the line between $(x,y)$ and $(3,8)$.

Step 2: Compute the slope of the tangent line to $f(x)=x^2$ at the point $(x,y)$.

Step 3: Set these two slopes equal to each other and find candidate $x$ values.

Step 4: Check your answers.

Answer 2

A point on the graph is $(x,x^2)$. The slope from that point to $(3,8)$ is given by: $\frac{x^2-8}{x-3}$. This has to be equal to the derivative at the point for it to be a tangent. So: $$\frac{x^2-8}{x-3}=2x$$ $$x^2-8=2x^2-6x$$ $$0=x^2-6x+8$$ $$0=(x-2)(x-4)$$ So $x=2$ or $x=4$. So the points are $(2,4)$ and $(4,16)$.

Answer 3

Either you write the equation of a tangent at $(x_0,x_0^)$2 and check under which condition this equation is satisfied by the $(3,8)$.

Or you write the equation of a line with slope $t$ passing through the point $(3,8)$: $$y-8=t(x-3),$$ and find under which condition this line has a double intersection with the parabola: $$x^2-8=t(x-3)\iff x^2 -tx+3t-8=0$$ One has a double root if and only if its discriminant: $$\Delta=t^2-12t+32=(t-6)^2-4=0\iff t=\begin{cases}4\\8\end{cases}$$ Th abscissée ot the points of contact are the double roots: $\dfrac t2=2,4$, and their ordinates $4,16$.