Question

If (2,-8) is one end of a focal chord of the parabola $y^2 = 32x$, then the other end of the focal chord, is

• A. (32,32)
• B. (32,-32)
• C. (-2,8)
• D. (2,8)

Answer 1

Answer: (A)

(32,32)

Answer 2

The given parabola is $y^2 = 32x$. This is in the form $y^2 = 4ax$, so $4a = 32$, which gives $a = 8$. The parametric form of a point on this parabola is $(at^2, 2at) = (8t^2, 16t)$. Let the given end of the focal chord be $P_1 = (2, -8)$. We can find the parameter $t_1$ for this point: $16t_1 = -8 \implies t_1 = -1/2$. $8t_1^2 = 8(-1/2)^2 = 8(1/4) = 2$. This matches the x-coordinate. For a focal chord, if one end corresponds to parameter $t_1$, the other end corresponds to parameter $t_2$ such that $t_1 t_2 = -1$. So, $(-1/2) t_2 = -1 \implies t_2 = 2$. The other end of the focal chord $P_2$ has coordinates $(8t_2^2, 16t_2)$. $P_2 = (8(2)^2, 16(2)) = (8 \times 4, 32) = (32, 32)$.