# Find all points $(x,y)$ on the graph of $f(x)$ with tangent lines passing through a certain point

Find all points $(x,y)$ on the graph of $f(x) = x^2$ with tangent lines passing through the point $(3,8)$.

My attempt:

$f'(x) = x^2$ I substituted a random point into $f'(x)$ to get the gradient of the tangent, so $f'(0) = 0$.

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$ $$m = \frac{8-0}{3-0} = 8/3$$

$$y-y_1 = m(x-x_1)$$ Now I substituted the point $(3,8)$: $$y - 8 = \frac{8}{3}(x-3)$$ $$y = \frac{8}{3}x$$ $$y = x^2$$

Solving simultaneously we get $x = 0$ and $y = 0$ or $x = 8/3$ and $y = 64/9$

So we have the points $(0;0)$ and $(\frac{8}{3} ; \frac{64}{9})$. However, I am not so confident with my answer. Is this correct?

Let $(a, a^2)$ be a point on the graph of $f(x) = x^2$ the tangent line at which passes through the point $(3, 8)$. We know that the slope of the tangent line at $(a, a^2)$ is $$\frac{\mathrm{d} f(x) }{\mathrm{d} x} \vert_{x=a} = 2a.$$ But the tangent line passes through the points $(a, a^2)$ and $(3, 8)$. So its slope is $$\frac{a^2 - 8}{a-3}.$$ Now equating the above two expressions for the slope, we obtain $$\frac{a^2 - 8}{a-3} = 2a.$$ which yields $$a^2 - 6a + 8 = 0,$$ from which we obtain $a = 2$ or $a=4$.

Hence the two points are $(2, 4)$ and $(4, 16)$.

$y = x^2\\ y' = 2x$

let $(x_0,y_0)$ be a point on the curve, and the line tangent to that point:

$(y-y_0) = 2x_0(x-x_0)\\ y-x_0^2 = 2x_0x-2x_0^2\\ y = 2x_0x - x_0^2$

Goes through the point $(3,8)$

$8 = 6x_0 - x_0^2\\ (x_0 - 2)(x_0 -4) = 0$

$(2,4), (4,16)$ are points on the curve where the line tangent to those points goes through the point $(3,8)$

Here is an alternative method:

A line with gradient $m$ passing through the point $(3,8)$ has equation $$y-8=m(x-3)$$ Solving this simultaneously with the curve $y=x^2$ leads to the quadratic equation $$x^2-mx+3m-8=0$$ This must have double roots so the discriminant is zero, so $$m^2-12m+32=0\implies m=4,8$$ Then $x=-\frac{b}{2a}=\frac m2=2,4$

So the points are $$(2,4),(4,16)$$