Question

Let O be the vertex of the parabola \(y^2 = 4x\) and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is:

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use the parametric coordinates of points on a parabola. Since the chords from the vertex are perpendicular, the product of their slopes must be $-1$. This condition allows us to find a relationship between the parameters, which we then use to find the locus of the midpoint.
Step 2: Key Formula or Approach:
1. Parametric point on \(y^2 = 4ax\): \((at^2, 2at)\). Here \(a=1\).
2. Slope \(m = \frac{2t - 0}{t^2 - 0} = \frac{2}{t}\).
3. For \(OP \perp OQ\): \(m_P \cdot m_Q = -1\).
Step 3: Detailed Explanation:
Let \(P = (t_1^2, 2t_1)\) and \(Q = (t_2^2, 2t_2)\). Slopes are \(m_P = \frac{2}{t_1}\) and \(m_Q = \frac{2}{t_2}\). Condition: \(\frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies t_1 t_2 = -4\). Midpoint \((h, k)\): \(h = \frac{t_1^2 + t_2^2}{2} = \frac{(t_1 + t_2)^2 - 2t_1 t_2}{2} = \frac{(t_1 + t_2)^2 + 8}{2}\) \(k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2\) Substitute \(t_1 + t_2 = k\) into the equation for \(h\): \(2h = k^2 + 8 \implies y^2 = 2x - 8 \implies y^2 = 2(x - 4)\). This is a parabola of the form \(Y^2 = 4AX\), where \(4A = 2\).
Step 4: Final Answer:
The length of the latus rectum is 2.