Question

Two perpendicular straight lines through the focus of the parabola $y^2=4ax$ meet its directrix in $T$ and $T^{\prime }$ respectively. If the tangents to the parabola parallel to the perpendicular lines intersect in the mid point of ${TT}^{\prime }.$.Find the coordinate of mid point.

• A. $[-2a,a(\frac{1}{m}+m)]$
• B. $[-a,a(\frac{2}{m}-m)]$
• C. $[-2a,a(\frac{2}{m}+m)]$
• D. $[-a,a(\frac{1}{m}-m)]$

Answer 1

The correct option is D $[-a,a(\frac{1}{m}-m)]$

Any line through the focus $(a,0)$ is
$y=m(x-a)...\left(1\right)$
Any line perpendicular to above through $(a,0)$ is
$y=-\frac{1}{m}[x-a)...\left(2\right)$
Solving the directrix x=-a, we get the points T and $T^{\prime }$
$T(-a,-2am)$ and $T^{\prime }(-a,\frac{2a}{m})$
Mid - point of ${TT}^{\prime }$ is $[-a,a(\frac{1}{m}-m)]...\left(3\right)$
Now tangents parallel to lines (1) and (2) are
$y=mx+\frac{a}{m},y=-\frac{1}{m}x-am.$
Substracting we get $0=x(m+\frac{1}{m})+a(m+\frac{1}{m})$
$\therefore x=-a,$ and hence on putting for x, we get $y=a(\frac{1}{m}-m).$
Thus the point of intersection is $[-a,a(\frac{1}{m}-m)]$
which is mid - point of ${TT}^{\prime }$ by (3).