Question

Let the locus of the mid-point of the chord through the origin \(O\) of the parabola \(y^2 = 4x\) be the curve \(S\). Let \(P\) be any point on \(S\). Then the locus of the point, which internally divides \(OP\) in the ratio \(3:1\), is

• A. \(3y^2 = 2x\)
• B. \(3x^2 = 2y\)
• C. \(2y^2 = 3x\)
• D. \(2x^2 = 3y\)

Hint:

Use parametric form and section formula together to find loci involving division of line segments.

Answer 1

The Correct Option is C

The given parabola is \[ y^2 = 4x. \]
Step 1: Equation of chord through the origin.
Any chord of the parabola passing through the origin has equation \[ y = mx. \] Substituting in the parabola equation, \[ m^2x^2 = 4x \Rightarrow x = \frac{4}{m^2}. \] Thus, the second point of intersection is \[ \left(\frac{4}{m^2}, \frac{4}{m}\right). \]
Step 2: Mid-point of the chord.
Let \(P(h,k)\) be the mid-point of the chord joining the origin and this point. Then, \[ h = \frac{2}{m^2}, \quad k = \frac{2}{m}. \] Eliminating \(m\), \[ k^2 = 2h. \] Hence, the locus \(S\) of the mid-point is \[ y^2 = 2x. \]
Step 3: Point dividing \(OP\) in the ratio \(3:1\).
Let \(Q(x,y)\) be the point dividing \(OP\) internally in the ratio \(3:1\). Using section formula, \[ x = \frac{3h}{4}, \quad y = \frac{3k}{4}. \] Substituting \(h = \frac{4x}{3}\) and \(k = \frac{4y}{3}\) into \(k^2 = 2h\), \[ \left(\frac{4y}{3}\right)^2 = 2\left(\frac{4x}{3}\right). \] \[ \frac{16y^2}{9} = \frac{8x}{3}. \] \[ 2y^2 = 3x. \]
Final Answer: \[ \boxed{2y^2 = 3x} \]