Question:
Through the vertex $O$ of a parabola $y^2 = 4x$, chords $OP$ and $OQ$ are drawn at right angles to one another. The locus of the middle point of $PQ$ is
Through the vertex $O$ of a parabola $y^2 = 4x$, chords $OP$ and $OQ$ are drawn at right angles to one another. The locus of the middle point of $PQ$ is
Updated On: Jan 30, 2025
- $y^2 = 2x + 8 $
- $y^2 = x + 8 $
- $y^2 = 2x - 8 $
- $y^2 = x - 8 $
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The Correct Option is C
Solution and Explanation
Given parabola is $y^2 = 4x$ ....(1)
Let $P \equiv\left(t^{2}_{1} , 2t_{1}\right)$ and $ Q \equiv \left(t^{2}_{2}, 2t_{2}\right) $
Slope of $ OP = \frac{2t_{1}}{t_{1}^{2}} = \frac{2}{t_{1}} $
and slope of $ OQ = \frac{2}{t_{2}} $
Since $ OP \bot OQ$,
$ \therefore \frac{4}{t_{1}t_{2}} = - 1 t_{1 }t_{2} = - 4 $ ....(2)
Let $R (h, k)$ be the middle point of $PQ$, then
$h = \frac{t_{1}^{2} + t^{2}_{2}}{ 2}$ ....(3)
and $ k = t_{1} +t_{2} $ ....(4)
From (4) , $ k^{2} = t^{2}_{1} + t^{2}_{2} + 2t_{1}t_{2} = 2h-8$
[ From (2) and (3)]
Hence locus of $R (h, k) $ is $y^2 = 2x - 8$.
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.