Q.11. Let O be the vertex of the parabola 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 rec

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Q.11. Let O be the vertex of the parabola  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 :

FirstInTest · Accepted Answer

Correct answer: B. The vertex of the parabola $y^2 = 4x$ is O(0,0).
Let P and Q be two points on the parabola. The parametric coordinates of a point on $y^2 = 4x$ are $(t^2, 2t)$.
Let $P = (t_1^2, 2t_1)$ and $Q = (t_2^2, 2t_2)$.

The slope of OP is $m_{OP} = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1}$.
The slope of OQ is $m_{OQ} = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2}{t_2}$.

Since OP and OQ are perpendicular, $m_{OP} \cdot m_{OQ} = -1$.
$(\frac{2}{t_1}) (\frac{2}{t_2}) = -1 \Rightarrow \frac{4}{t_1 t_2} = -1 \Rightarrow t_1 t_2 = -4$.

Let $(h, k)$ be the mid-point of PQ.
$h = \frac{t_1^2 + t_2^2}{2}$ and $k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$.

We know that $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1 t_2$.
Substitute $k = t_1+t_2$ and $t_1 t_2 = -4$ into this equation:
$2h = k^2 - 2(-4) = k^2 + 8$.

So, the locus of the mid-point $(h, k)$ is $2x = y^2 + 8$, or $y^2 = 2x - 8$.
This is a parabola. To find its latus rectum, we can rewrite it in the standard form $Y^2 = 4AX$.
$y^2 = 2(x - 4)$.
Comparing with $Y^2 = 4AX$, we have $Y=y$, $X=x-4$, and $4A=2$.
So, $A = 1/2$.
The length of the latus rectum is $4A = 2$.

The conic C is a parabola with latus rectum length 2.

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