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

• A. \(41\)
• B. \(89\)
• C. \(32\)
• D. \(80\)

Hint:

For chord problems in parabolas, centroid conditions often directly give sum of coordinates.

Answer 1

The Correct Option is C

Step 1: Coordinates of focus For \(y^2=4ax\), focus is \((a,0)\). Here \(4a=8\Rightarrow a=2\). \[ A=(2,0) \]
Step 2: Let points of intersection be \(B(x_1,y_1)\) and \(C(x_2,y_2)\) They satisfy: \[ y=mx+c,\quad y^2=8x \] Using centroid formula: \[ \left(\frac{x_1+x_2+2}{3},\frac{y_1+y_2}{3}\right)=\left(\frac{7}{3},\frac{4}{3}\right) \] Thus: \[ x_1+x_2=5,\quad y_1+y_2=4 \]
Step 3: Distance \(BC\) \[ (BC)^2=(x_1-x_2)^2+(y_1-y_2)^2 \] Using identities: \[ (x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2 \] \[ (y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2 \] From parabola relation: \[ y^2=8x \Rightarrow y_1^2+y_2^2=8(x_1+x_2)=40 \] Also, \[ (y_1+y_2)^2=y_1^2+y_2^2+2y_1y_2 \Rightarrow 16=40+2y_1y_2 \Rightarrow y_1y_2=-12 \] Similarly, \[ x_1x_2=\frac{y_1^2y_2^2}{64}=\frac{144}{64}=\frac{9}{4} \]
Step 4: Compute \((BC)^2\) \[ (BC)^2=(25-9)+(16-4(-12))=16+16=32 \] Final Answer: \[ \boxed{32} \]