Question

If from the vertex of the parabola $y^2=4ax$ pair of chords be drawn perpendicular to each other and with these chords as adjacent sides a rectangle is completed then the locus of the vertex of the farther angle of the rectangle is the parabola

• A. $y^2=4a(x-4a)$
• B. $y^2=4a(x-6a)$
• C. $y^2=4a(x-2a)$
• D. $y^2=4a(x-8a)$

Answer 1

The correct option is D $y^2=4a(x-8a)$

$OA\perp OB\Rightarrow t_1{t}_2=-4$
if $C$ is $(h,k),,$ then the diagonals of a rectangle bisect each other.
$h+0=a(t_1^2+t_2^2)$
$k+0=2a(t_1+t_2)$
$\therefore h=a\left[\right(t_1+t_2{)}^2-2t_1{t}_2]=a[\frac{k^2}{4a^2}+8]$
or
$(h-8a)=\frac{k^2}{4a}$
$\therefore$ Locus is $y^2=4a(x-8a)$