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 :
Solutions · 1
Official solution
Given, equation of parabola \[\Rightarrow\] y2 = 4ax
Focus = S(a, 0)
Let any point on the parabola be P(at2, 2at).
and let the mid-point of PS be M(h, k).
\[\therefore\] \[h{{a{t^2} + a} \over 2};k = {{2at + 0} \over 2}\]
\[ \Rightarrow {t^2} = {{2h - a} \over a};t = {k \over a}\]
\[ \Rightarrow {t^2} = {{{k^2}} \over {{a^2}}}\]
Now, \[{{2h - a} \over a} = {{{k^2}} \over {{a^2}}}\]
\[ \Rightarrow 2h - a = {{{k^2}} \over a} \Rightarrow {k^2} = a(2h - a)\]
\[\therefore\] Locus of (h, k) is y2=a(2x−a)
\[{y^2} = 2a\left( {x - {a \over 2}} \right)\]
\[\therefore\] The directrix of this parabola is
\[x - {a \over 2} = - {a \over 2} \Rightarrow x = 0\]