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

Accepted Answer

Identify the Parabola Parameters The parabola is ${y}^{2}=8x$. Comparing this with the standard form ${y}^{2}=4ax$, we find $4a=8$, so $a=2$.

Focus $A$ : The focus is $(a,0)$, which is $(2,0)$.

Define Parametric Points for $B$ and $C$ : Let the two distinct points of intersection be $B$ a $C$. Using parametric coordinates $(a{t}^{2},2at)$ :

Point $B=(2{t}_{2}^{1},4{t}_{1})$

Use the Centroid Formula : The centroid $G(x,y)$ of a triangle with vertices $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$, and $({x}_{3},{y}_{3})$ is:

$G=(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3})$

Use the Centroid Property : The centroid of triangle $ABC$ with vertices $A(2,0)$, $B(2{t}_{2}^{1},4{t}_{1})$, and $C(2{t}_{2}^{2},4{t}_{2})$ is given as $(\frac{7}{3},\frac{4}{3})$.

For the $y$-coordinate :

$\frac{4{t}_{1}+4{t}_{2}+0}{3}=\frac{4}{3}\implies 4({t}_{1}+{t}_{2})=4\implies {t}_{1}+{t}_{2}=1$

For the $x$-coordinate :

$\frac{2{t}_{2}^{1}+2{t}_{2}^{2}+2}{3}=\frac{7}{3}\implies 2({t}_{2}^{1}+{t}_{2}^{2})=5\implies {t}_{2}^{1}+{t}_{2}^{2}=\frac{5}{2}$

Find the Relationship between ${t}_{1}$ and ${t}_{2}$ :

Using the identity ${({t}_{1}+{t}_{2})}^{2}={t}_{2}^{1}+{t}_{2}^{2}+2{t}_{1}{t}_{2}$ :

$(1{)}^{2}=\frac{5}{2}+2{t}_{1}{t}_{2}\implies 1-\frac{5}{2}=2{t}_{1}{t}_{2}\implies 2{t}_{1}{t}_{2}=-\frac{3}{2}\implies {t}_{1}{t}_{2}=-\frac{3}{4}$

Now, calculate ${({t}_{1}-{t}_{2})}^{2}$ :

${({t}_{1}-{t}_{2})}^{2}={({t}_{1}+{t}_{2})}^{2}-4{t}_{1}{t}_{2}={1}^{2}-4(-\frac{3}{4})=1+3=4$

Calculate $(BC{)}^{2}$ The distance formula for $(BC{)}^{2}$ is :

$(BC{)}^{2}={({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}$

$\Rightarrow$ $(BC{)}^{2}={[2({t}_{2}^{2}-{t}_{2}^{1})]}^{2}+{[4({t}_{2}-{t}_{1})]}^{2}$

$\Rightarrow$ $(BC{)}^{2}=4{({t}_{2}-{t}_{1})}^{2}{({t}_{2}+{t}_{1})}^{2}+16{({t}_{2}-{t}_{1})}^{2}$

Substitute the values ${({t}_{1}-{t}_{2})}^{2}=4$ and $({t}_{1}+{t}_{2})=1$ :

$(BC{)}^{2}=4(4)(1{)}^{2}+16(4)=16+64=80$

Solutions · 1