Question

The locus of a point which divides the line segment joining the focus and any point on the parabola \(y^2 = 12x\) in the ratio \(m:n\) (\(m+n \ne 0\)) is a parabola. Then the length of the latus rectum of that parabola is

• A. \(\frac{m}{m+n}\)
• B. \(\frac{12m}{m+n}\)
• C. \(\frac{m}{12(m+n)}\)
• D. \(\frac{n}{12(m+n)}\)

Hint:

When finding a locus involving a parameter (like \(t\)), express the coordinates \((h,k)\) in terms of \(t\), then solve one equation for \(t\) and substitute into the other to eliminate it.

Answer 1

The Correct Option is B

Step 1: Coordinates of Focus and Point on Parabola:
Parabola: \(y^2 = 12x \implies 4a = 12 \implies a = 3\). Focus \(S(3, 0)\). Let \(P(at^2, 2at) = (3t^2, 6t)\) be any point on the parabola.
Step 2: Apply Section Formula:
Let \(Q(h, k)\) be the point dividing \(SP\) in the ratio \(m:n\). \[ h = \frac{m(x_P) + n(x_S)}{m+n} = \frac{m(3t^2) + n(3)}{m+n} \] \[ k = \frac{m(y_P) + n(y_S)}{m+n} = \frac{m(6t) + n(0)}{m+n} = \frac{6mt}{m+n} \]
Step 3: Eliminate Parameter t:
From the k-equation: \(t = \frac{k(m+n)}{6m}\). Substitute into the h-equation: \[ h = \frac{3m}{m+n} \left( \frac{k(m+n)}{6m} \right)^2 + \frac{3n}{m+n} \] \[ h - \frac{3n}{m+n} = \frac{3m}{m+n} \cdot \frac{k^2 (m+n)^2}{36m^2} \] \[ h - \frac{3n}{m+n} = \frac{k^2 (m+n)}{12m} \] Rearranging to form \(Y^2 = 4AX\): \[ k^2 = \frac{12m}{m+n} \left( h - \frac{3n}{m+n} \right) \]
Step 4: Identify Latus Rectum:
The equation is of the form \(y^2 = 4A(x - x_0)\). The length of the latus rectum is the coefficient \(4A\). \[ \text{Length of LR} = \frac{12m}{m+n} \]