Question

Locus of midpoints of all chords of the parabola ${y^2} = 4x$ which are drawn through the vertex is
A.${y^2} = 8x$
B.${y^2} = 2x$
C.${x^2} + 4{y^2} = 16$
D.${x^2} = 2y$

Answer 1

Hint: The two points of the chord are the vertex of parabola and the other end is on parabola itself. And a point is midpoint on this chord only. First we will find coordinates of the midpoint and then substituting them with the ordinate and abscissa of midpoint only we will get the locus of the midpoints.

Complete step-by-step answer:

First we will observe the diagram of the parabola given below.

Here chord is OQ.

The two end points of chords are one is vertex of parabola that is origin of coordinate system $O\left( {0,0} \right)$ and other is $Q\left( {a{t^2},2at} \right)$. The midpoint of the chord is$P\left( {x,y} \right)$.

Now using midpoint formula,

$\Rightarrow P\left( {\dfrac{{a{t^2} + 0}}{2},\dfrac{{2at + 0}}{2}} \right)$

$\Rightarrow P\left( {\dfrac{{a{t^2}}}{2},\dfrac{{2at}}{2}} \right)$

$\Rightarrow P\left( {\dfrac{{a{t^2}}}{2},at} \right)$

But assuming

$x = \dfrac{{a{t^2}}}{2}$ and $y = at$

Squaring the y value

${y^2} = {a^2}{t^2}$

Putting the value of ${t^2}$ from value of x,

$\Rightarrow {y^2} = \dfrac{{{a^2}2x}}{a}$

$\Rightarrow {y^2} = 2ax$

Since the value of a is 1 so,

$\Rightarrow {y^2} = 2x$

And this is locus of midpoints of all chords of the parabola which are drawn through the vertex.

Thus option B is the correct option.

Note: Here chord is drawn through the vertex so one of the ends of the chord is necessarily the origin of the coordinate system. Don’t forget to put the value of a is 1.