Question

Let \(O\) be the vertex of the parabola \(y^2 = 4x\). Let \(P\) and \(Q\) be two points on parabola such that chords \(OP\) and \(OQ\) are perpendicular to each other. If the locus of mid-point of segment \(PQ\) is a conic \(C\), then latus rectum of \(C\) is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

Step 1: Parametric Representation:
A general point on $y^2 = 4x$ can be written as $(t^2, 2t)$ using the standard parametrization.
Let $P = (t_1^2, 2t_1)$ and $Q = (t_2^2, 2t_2)$, and $O = (0,0)$ (the vertex).
\includegraphics[width=0.5\linewidth]{m13.png}
Step 2: Perpendicularity Condition:
Slope of $OP$: \[ m_1 = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1} \] Slope of $OQ$: \[ m_2 = \frac{2}{t_2} \] For perpendicularity: $m_1 \cdot m_2 = -1$: \[ \frac{2}{t_1}\cdot\frac{2}{t_2} = -1 \implies t_1 t_2 = -4 \] Step 3: Finding the Locus of the Midpoint:
Let $R = (h, k)$ be the midpoint of $PQ$: \[ h = \frac{t_1^2 + t_2^2}{2}, \qquad k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2 \] Now use the identity $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1t_2$: \[ 2h = (t_1+t_2)^2 - 2t_1t_2 = k^2 - 2(-4) = k^2 + 8 \] \[ k^2 = 2h - 8 = 2(h - 4) \] Step 4: Identifying the Conic and its Latus Rectum:
Replacing $(h,k)$ by $(x,y)$: $y^2 = 2(x-4)$.
This is a parabola with vertex at $(4,0)$ and $4a = 2$, so $a = \dfrac{1}{2}$.
The latus rectum of a parabola $y^2 = 4ax$ is $4a$.
Here $4a = 2$, so the latus rectum $= 2$.
Step 5: Final Answer:
The latus rectum of conic $C$ is $\mathbf{2}$.
The answer is Option (2).