JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 12)
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 :
x = 0
x = - $${a \over 2}$$
x = a
x = $${a \over 2}$$
Explanation
Given, equation of parabola $$\Rightarrow$$ y2 = 4ax
Focus = S(a, 0)
Let any point on the parabola be P(at2, 2at).
_24th_February_Morning_Shift_en_12_1.png)
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 $${y^2} = a(2x - a)$$
$${y^2} = 2a\left( {x - {a \over 2}} \right)$$
$$\therefore$$ The directrix of this parabola is
$$x - {a \over 2} = - {a \over 2} \Rightarrow x = 0$$
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 $${y^2} = a(2x - a)$$
$${y^2} = 2a\left( {x - {a \over 2}} \right)$$
$$\therefore$$ The directrix of this parabola is
$$x - {a \over 2} = - {a \over 2} \Rightarrow x = 0$$
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