<s>[INST] 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?
 [/INST] 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 $$(2,4),(4,16)$$
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