Let A be the focus of the parabola y^2=8x.

Question

Let $A$ be the focus of the parabola $y^2=8x$. Let the line $y=mx+c$ intersect the parabola at two distinct points $B$ and $C$. If the centroid of the triangle $ABC$ is $\left(\dfrac73,\dfrac43\right)$, then $(BC)^2$ is equal to:

Accepted Answer

Here $A=(2,0)$, and points on the parabola are $(2t^2,4t)$. Let $B,C$ correspond to $t_1,t_2$.

$$\frac{2+2t_1^2+2t_2^2}{3}=\frac73 \Rightarrow t_1^2+t_2^2=\frac52,\qquad \frac{4t_1+4t_2}{3}=\frac43\Rightarrow t_1+t_2=1.$$

$$2t_1t_2=(t_1+t_2)^2-(t_1^2+t_2^2)=1-\frac52=-\frac32 \Rightarrow t_1t_2=-\frac34.$$

$(t_1-t_2)^2=(t_1+t_2)^2-4t_1t_2=1+3=4.$

$$(BC)^2=4(t_1^2-t_2^2)^2+16(t_1-t_2)^2 = 4\big[(t_1-t_2)(t_1+t_2)\big]^2+16(4)=4(4)+64=80.$$

Solutions · 1