Question

Find the locus of the middle points of chords of the parabola which are of given length l.

Answer 1

For the parabola $y^2=4ax$, chord joining points $(a{t}_1^2,2a{t}_1)$ and $(a{t}_2^2,2a{t}_2)$ has the equation $y-2a{t}_2=\frac{2a{t}_2-2a{t}_1}{a{t}_2^2-a{t}_1^2}\times (x-a{t}_2^2)$

i.e. $(y-2a{t}_2)(t_1+t_2)=2(x-a{t}_2^2)$

The chords are of length $l$

$\Rightarrow (a{t}_2^2-a{t}_1^2{)}^2+(2a{t}_2-2a{t}_1{)}^2=l^2$

$\Rightarrow a^2(t_1-t_2{)}^2[(t_1+t_2{)}^2+4]=l^2...\left(1\right)$

The midpoint of the chord is given by $(\frac{a(t_1^2+t_2^2)}{2},a(t_1+t_2\left)\right)$

Let the midpoint be denoted by $(x,y)$

$\therefore 2x=a(t_1^2+t_2^2),y=a(t_1+t_2)$

Equation (1) becomes $a^2\times [\frac{y^2}{a^2}-4(\frac{y^2}{2a^2}-\frac{2x}{2a})]\times (\frac{y^2}{a^2}+4)=l^2$

$\Rightarrow (y^2-2y^2+4ax)(y^2+4a^2)=a^2{l}^2$

$\Rightarrow (4ax-y^2)(y^2+4a^2)=a^2{l}^2$

i.e. $y^4+y^2(4a^2-4ax)-16a^{3}x+a^2{l}^2=0$ is the required locus