Question

If from the vertex of a parabola $y^2 = 4ax$ a pair of chords be drawn at right angles to one another and with these chords as adjacent sides a rectangle be made, then the locus of the further angle of the rectangle is:

• A. an equal parabola
• B. a parabola with focus at (8a, 0)
• C. a parabola with directrix as $x-7a=0$
• D. not a parabola

Answer 1

Answer: (A), (C)

a parabola with directrix as $x-7a=0$, an equal parabola

Answer 2

Let the parabola be $y^2 = 4ax$. The vertex is $O(0,0)$. Let the two chords from the vertex be $OA$ and $OB$, where $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$. The slopes of $OA$ and $OB$ are $m_{OA} = 2/t_1$ and $m_{OB} = 2/t_2$. Since $OA \perp OB$, $m_{OA} \cdot m_{OB} = -1$, so $(2/t_1)(2/t_2) = -1$, which gives $t_1t_2 = -4$. Let $C(h,k)$ be the fourth vertex of the rectangle $OACB$. Equating the midpoints of diagonals $OC$ and $AB$: $h = a(t_1^2+t_2^2)$ and $k = 2a(t_1+t_2)$. Using $(t_1+t_2)^2 = t_1^2+t_2^2 + 2t_1t_2$, we substitute $t_1+t_2 = k/2a$, $t_1^2+t_2^2 = h/a$, and $t_1t_2 = -4$. This yields $(k/2a)^2 = h/a + 2(-4)$, simplifying to $k^2/(4a^2) = h/a - 8$. Rearranging, $k^2 = 4a^2(h/a - 8) = 4a(h - 8a)$. The locus of $C(h,k)$ is $y^2 = 4a(x - 8a)$.

This is a parabola with the same parameter '$a$' as the original parabola, hence it is an "equal parabola". The focus of $y^2 = 4a(x - 8a)$ is at $(8a+a, 0) = (9a, 0)$. The directrix of $y^2 = 4a(x - 8a)$ is $x - 8a = -a$, which simplifies to $x = 7a$, or $x-7a=0$. Therefore, options (a) and (c) are correct.