Let A be the focus of the parabola y^{2}=8x. | vedclass

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(B) The parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$. The focus $A$ is $(2,0)$.Let the points $B$ and $C$ be $(2t_{1}^{2}, 4t_{1})$ and $(2t_{2}^{

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$80$

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(B) The parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$. The focus $A$ is $(2,0)$.Let the points $B$ and $C$ be $(2t_{1}^{2}, 4t_{1})$ and $(2t_{2}^{2}, 4t_{2})$ respectively.The centroid of $	riangle ABC$ is given by $(\frac{2t_{1}^{2}+2t_{2}^{2}+2}{3}, \frac{4t_{1}+4t_{2}+0}{3}) = (\frac{7}{3}, \frac{4}{3})$.Equating the coordinates:$2(t_{1}^{2}+t_{2}^{2}+1) = 7 \Rightarrow t_{1}^{2}+t_{2}^{2} = \frac{5}{2}$.$4(t_{1}+t_{2}) = 4 \Rightarrow t_{1}+t_{2} = 1$.Now,$(t_{1}-t_{2})^{2} = (t_{1}+t_{2})^{2} - 4t_{1}t_{2}$.Since $t_{1}^{2}+t_{2}^{2} = (t_{1}+t_{2})^{2} - 2t_{1}t_{2} = \frac{5}{2}$,we have $1 - 2t_{1}t_{2} = \frac{5}{2}$ $\Rightarrow 2t_{1}t_{2} = -\frac{3}{2}$ $\Rightarrow t_{1}t_{2} = -\frac{3}{4}$.Thus,$(t_{1}-t_{2})^{2} = 1 - 4(-\frac{3}{4}) = 1+3 = 4$.$(BC)^{2} = (2t_{1}^{2}-2t_{2}^{2})^{2} + (4t_{1}-4t_{2})^{2} = 4(t_{1}^{2}-t_{2}^{2})^{2} + 16(t_{1}-t_{2})^{2}$.$(BC)^{2} = 4(t_{1}+t_{2})^{2}(t_{1}-t_{2})^{2} + 16(t_{1}-t_{2})^{2} = 4(1)^{2}(4) + 16(4) = 16 + 64 = 80$.

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 $(\frac{7}{3},\frac{4}{3})$,then $(BC)^{2}$ is equal to:

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