Question

A parabola touches two given straight lines at given points; prove that the locus of the middle point of the portion of any tangent which is intercepted between the given straight lines is a straight line.

Answer 1

Point of intersection of the tangents drawn at the ends of a chord of $y^2=4ax$ is $R=(a{t}_1{t}_2,a(t_1+t_2\left)\right)$

Let the two fixed point of contact be $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$

And the point of contact of variable tangent be $A(a{t}^2,2at)$ and the mid point of portion be $(h,k)$

Point of intersection of tangent at $P$ and $A$ is $B(at{t}_1,a(t+t_1\left)\right)$

Point of intersection of tangent at $Q$ and $A$ is $C(at{t}_2,a(t+t_2\left)\right)$

Mid point of $BC$ is

$(\frac{at{t}_1+at{t}_2}{2},\frac{a(t+t_1)+a(t+t_2)}{2})(\frac{at(t_1+t_2)}{2},\frac{2at+a(t_1+t_2)}{2})\frac{at(t_1+t_2)}{2}=h\Rightarrow at=\frac{2h}{t_1+t_2}.............\left(i\right)\frac{2at+a(t_1+t_2)}{2}=k2at+a(t_1+t_2)=2k$

Substituting $at$ from $\left(i\right)$

$\frac{4h}{t_1+t_2}+a(t_1+t_2)=2k4h-2k(t_1+t_2)+a{{(t}_1+t_2)}^2=04x-2y(t_1+t_2)+a{{(t}_1+t_2)}^2=0$

As $t_1$ and $t_1$ are fixed therefore the equation represents a straight line

Hence proved.