# Find the equation of the tangent to the parabola $y=x^2$, if the x-intercept of the tangent is 2

I'm trying to solve this problem:

Find the equation of the tangent to the parabola $y=x^2$.

If the x-intercept of the tangent is 2.

All what I can think of is finding the slope which is

$dy/dx = 2x$

so the tangent line equation would be

$y-y_0 = 2x(x-x_0)$

I don't know exactly where to go from here, should I plug in intercept points?

$y-0 = 2x(x-2)$

• You cannot do in this way. The $x$ in $dy/dx=2x$ is referring to the parabola. The $x$ in $y-y_0=m(x-x_0)$ is referring to the tangent. You cannot mix them up. Try to let the slope be $m$ and solve it by plugging in the intercept. Then use $dy/dx=2x=m$ to find where does the tangent cut the parabola. – Brian Cheung May 9 '16 at 17:51
• @user3313320 So it'll be $0-y_0 = m(2-x_0)$ ? – Rafael Adel May 9 '16 at 18:00
• It should be $y-0=m(x-2)$ – Brian Cheung May 10 '16 at 8:44

let the point where tangent is drawn to curve be $(a,b)$

so slope of tangent at that point is $\mathrm{\dfrac{dy}{dx}}=2x=2a$

so the equation of tangent at point $(a,b)$: $$y-b=2a(x-a)\,\,\,\,\,\,\,\,(3)$$ To find to find x-intercept we put $y=0$ (why? you can ask in comments)0-b=2a(x-a)\implies x=\frac{-b}{2a}+a=2\,\,\,\,\,\,\,\,\ (1)$$(because x- intercept i.e x=2)$$\implies b=2a^2-4a$$(a,b) lie on the parabola so they must satisfy the parabola equation y=x^2 so,$$b=a^2\,\,\,\,\,\,\,\,\,(2) on solving equation $1$ and $2$, we get two values of $a$ as $0,4$

now you know $a$, you can find $b$ then plug values in the equation $(3)$. And you will get two equations and that solves the problem :D

• diagrams are rough – user5954246 May 9 '16 at 18:09
• In equation (2) should it be $\frac {b-0} {a - 2} = 2a$ ? – Rafael Adel May 9 '16 at 18:26
• @RafaelAdel my bad!! i did wrong! let me edit my answer – user5954246 May 9 '16 at 18:28
• let me do on pen paper, sorry ;p – user5954246 May 9 '16 at 18:30
• Ah! Thank you so much. – Rafael Adel May 9 '16 at 19:09