If the line $y=x-8$ intersects the parabola ${ y }^{ 2 }=4x$ at A and B, what's the chord's length?

Question

How to find the coordinates of the points A and B? If one gets that, then distance formula can be used to find the length.

Answer 1

Let us denote the intersection of $y^2=4x$ and $y=x-8$ as $(a,b)$. Note that the coordinates $(a,b)$ satisfies $$b^2=4a=(a-8)^2 \iff a^2-20a+64=(a-4)(a-16)=0$$ As $(a-8)^2-4a=a^2-20a+64$. So the intersections are $(4, -4)$ and $(16, 8)$. So the distance between the two points is $12 \sqrt{2}$.

Answer 2

Any point on $y^2=4x$ can be set as $(t^2,2t)$

$x-y+8=0\implies t^2-2t-8=0\implies t_1,t_2$ are $4,-2$

Now we need $\sqrt{(t_1^2-t_2^2)^2+(2t_1-2t_2)^2}$