Question

From point \(B(4,8)\) on the parabola \(y^2 = 16x\), two perpendicular chords \(BA\) and \(BC\) are drawn. Given that the locus of the centroid of triangle \(BAC\) is another parabola with length of the latus rectum equal to \(\ell\), then \(3\ell\) is equal to

• A. \(14\)
• B. \(15\)
• C. \(12\)
• D. \(16\)

Hint:

For \(y^2=4ax\), parametric coordinates are \((at^2,2at)\). Using parameters simplifies problems involving chords and loci on parabolas.

Answer 1

The Correct Option is D

Concept: 

For the parabola \(y^2=4ax\), a general point can be written in parametric form as \[ (4at^2,\,4at) \] For \(y^2=16x\), \[ 4a=16 \Rightarrow a=4 \] Hence a general point on the parabola is \[ (4t^2,\,8t) \] 
Step 1: Take variable points on parabola Let \[ A(4t_1^2,\,8t_1), \quad C(4t_2^2,\,8t_2) \] and given point \[ B(4,8) \] 
Step 2: Condition of perpendicular chords Slopes of \(AB\) and \(BC\): \[ m_{AB}=\frac{8-8t_1}{4-4t_1^2}, \quad m_{BC}=\frac{8t_2-8}{4t_2^2-4} \] Since chords are perpendicular, \[ m_{AB} \cdot m_{BC} = -1 \] which simplifies to \[ t_1 + t_2 + t_1 t_2 = -5 \] 
Step 3: Centroid of triangle Centroid \(G(h,k)\) of triangle \(ABC\): \[ h=\frac{4t_1^2+4+4t_2^2}{3}, \qquad k=\frac{8t_1+8+8t_2}{3} \] \[ h=\frac{4(t_1^2+t_2^2+1)}{3}, \qquad k=\frac{8(t_1+t_2+1)}{3} \] 
Step 4: Eliminate parameters Using \[ t_1+t_2+t_1t_2=-5 \] and simplifying, the locus of centroid becomes \[ x=\frac{9}{48}y^2+\frac{40}{3} \] 
Step 5: Identify parabola parameters Comparing with standard form \[ x=\frac{1}{4a}y^2 + c \] \[ \frac{1}{4a}=\frac{9}{48} \] \[ 4a=\frac{48}{9} \] Hence length of latus rectum \[ \ell=\frac{48}{9} \] 
Step 6: Find \(3\ell\) \[ 3\ell = 3\left(\frac{48}{9}\right) \] \[ 3\ell = 16 \]