Question

Let \(\triangle ABC\) such that \(A(0,0)\) and vertices \(B\) and \(C\) lie on the parabola \[ y^2=8x. \] If \(\left(\frac{7}{3},\frac{4}{3}\right)\) is the centroid of \(\triangle ABC\), then \((BC)^2\) is equal to:

• A. \(110\)
• B. \(115\)
• C. \(120\)
• D. \(130\)

Hint:

When points lie on \(y^2=8x\), write \(x=\frac{y^2}{8}\). Using centroid relations helps convert geometry into simple algebra.

Answer 1

The Correct Option is C

Step 1: Use centroid formula
Let \[ B(x_1,y_1),\quad C(x_2,y_2). \] Given centroid \(G\): \[ G\left(\frac{x_1+x_2}{3},\frac{y_1+y_2}{3}\right) =\left(\frac{7}{3},\frac{4}{3}\right). \] Hence, \[ x_1+x_2=7,\qquad y_1+y_2=4. \]
Step 2: Use parabola condition
Since \(B\) and \(C\) lie on \(y^2=8x\), \[ x_1=\frac{y_1^2}{8},\quad x_2=\frac{y_2^2}{8}. \] Thus, \[ \frac{y_1^2+y_2^2}{8}=7 \;\Rightarrow\; y_1^2+y_2^2=56. \]
Step 3: Find \(y_1y_2\)
\[ (y_1+y_2)^2=y_1^2+y_2^2+2y_1y_2 \] \[ 4^2=56+2y_1y_2 \Rightarrow 16=56+2y_1y_2 \Rightarrow y_1y_2=-20. \]
Step 4: Compute \((BC)^2\)
\[ (BC)^2=(x_1-x_2)^2+(y_1-y_2)^2. \] First, \[ (y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2 =16-4(-20)=96. \] Next, \[ x_1-x_2=\frac{y_1^2-y_2^2}{8} =\frac{(y_1-y_2)(y_1+y_2)}{8} =\frac{4(y_1-y_2)}{8}=\frac{y_1-y_2}{2}. \] So, \[ (x_1-x_2)^2=\frac{(y_1-y_2)^2}{4} =\frac{96}{4}=24. \] Therefore, \[ (BC)^2=96+24=120. \] \[ \boxed{120} \]