Question

Find the locus of the middle points of chords of the parabola which pass through each point of the straight line $x=my+h$ is drawn the chord of the parabola $y^2=4ax$ which is bisected at the point; prove that it always touches the parahola
$(y+2am{)}^2=8a(x-h)$.

Answer 1

Let $P(x_1,y_1)$ be the mid point of the chords

Equation of chord when mid point is given is $T=S^{\prime }$

$y{y}_1-2a(x+x_1)=y_1^2-4a{x}_1.....\left(i\right)$

$P$ also lies on $x=my+h$

$x_1=m{y}_1+h$

Substituting in $\left(i\right)$, we get

$y{y}_1-2a(x+m{y}_1+h)=y_1^2-4a(m{y}_1+h)y{y}_1-2ax-2am{y}_1-2ah=y_1^2-4am{y}_1-4ahy{y}_1+2am{y}_1-2ax+2ah=y_1^2{y}_1(y+2am)-2a(x-h)=y_1^2(y+2am)=\frac{2a}{y_1}(x-h)+y_1$

$(y+2am)=\frac{2a}{y_1}(x-h)+\frac{2a}{\left(\frac{2a}{y_1}\right)}......\left(ii\right)$

Changing the origin to $(-2am,h)$ and putting $\frac{2a}{y_1}=M$ , equation $\left(ii\right)$ becomes

$Y=MX+\frac{2a}{M}$ , which is always a tangent to parabola

$Y^2=4\left(2a\right)X$

$Y^2=8aX.......\left(iii\right)$

with vertex $(-2am,h)$ the original equation become

${(y+2am)}^2=8a(x-h)$

Hence proved.