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

Updated On: Jun 25, 2024
  • $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).

Parabola


 

 

 

 

 

 

 

 

 

Standard Equation of a Parabola

For horizontal parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A,
  1. Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
  2. 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
  1. Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
  2. 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.