1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $(4, 16)$.

$\endgroup$
0
$\begingroup$

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

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 $(3,8)$

$\endgroup$
0
$\begingroup$

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 $$(2,4),(4,16)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.