Question

From the vertex of the parabola $y^2=4ax$ pair of perpendicular chords are drawn. If this chords are adjacent sides of a rectangle, then the locus of the vertex of the rectangle diagonally opposite to the vertex of parabola is

• 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 $y^2=4a(x-8a)$

Given parabola is $y^2=4ax$
Let the vertex be $O=(0,0)$ and the chords be $OA$ and $OB$
Where $A=(a{t}_1^2,2a{t}_1),B=(a{t}_2^2,2a{t}_2)$

Now, the slope of the line $OA$ is
$m_1=\frac{2}{t_1}$
Slope of the line $OB$ is
$m_2=\frac{2}{t_2}\Rightarrow m_1.m_2=-1(\because OA\perp OB)\Rightarrow t_1{t}_2=-4$

Let the coordinates of the vertex opposite to the vertex of parabola be $C=(h,k)$
As the diagonals of a rectangle bisect each other, so
$(\frac{h}{2},\frac{k}{2})=(\frac{a(t_1^2+t_2^2)}{2},\frac{2a(t_1+t_2)}{2})\Rightarrow \frac{h}{a}=t_1^2+\frac{16}{t_1^2},\frac{k}{2a}=t_1-\frac{4}{t_1}\Rightarrow {\left(\frac{k}{2a}\right)}^2=\frac{h}{a}-8$

Hence, the locus is
$y^2=4a(x-8a)$