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?
 A: 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 $(4, 16)$. 
A: $y = x^2\\
y' = 2x$
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 $(3,8)$
A: 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 $$(2,4),(4,16)$$
