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QJEEM21Ma8921: JEE Main 2021 Mathematics - Class 11 JEEM21Ma8921 Solution: The locus of the mid-point of the line segment joining the…

The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is :

Multiple Choice Options

  • A x = 0
  • B x = - a2{a \over 2}
  • C x = a
  • D x = a2{a \over 2}

Step-by-Step Verified Solution

Given, equation of parabola ⇒\Rightarrow y2 = 4ax

Focus = S(a, 0)

Let any point on the parabola be P(at2, 2at).

JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Parabola Question 82 English Explanation
and let the mid-point of PS be M(h, k).

∴\therefore hat2+a2;k=2at+02h{{a{t^2} + a} \over 2};k = {{2at + 0} \over 2}

⇒t2=2h−aa;t=ka\Rightarrow {t^2} = {{2h - a} \over a};t = {k \over a}

⇒t2=k2a2\Rightarrow {t^2} = {{{k^2}} \over {{a^2}}}

Now, 2h−aa=k2a2{{2h - a} \over a} = {{{k^2}} \over {{a^2}}}

⇒2h−a=k2a⇒k2=a(2h−a)\Rightarrow 2h - a = {{{k^2}} \over a} \Rightarrow {k^2} = a(2h - a)

∴\therefore Locus of (h, k) is y2=a(2x−a){y^2} = a(2x - a)

y2=2a(x−a2){y^2} = 2a\left( {x - {a \over 2}} \right)

∴\therefore The directrix of this parabola is

x−a2=−a2⇒x=0x - {a \over 2} = - {a \over 2} \Rightarrow x = 0

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