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?
Best Answer
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 $(4, 16)$.