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

Question

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?

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

Answer 1

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