Question

Find the coordinates of a point on the parabola $y^2 = 8x$ whose distance from the focus is 10.

Answer 1

(8, 8) and (8, -8)

Answer 2

The parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$, we get $4a = 8$, so $a = 2$. The focus is at $(2, 0)$. Let the point be $P(x, y)$. The distance from the focus to $P$ is given as 10.

Using the distance formula: $\sqrt{(x-2)^2 + (y-0)^2} = 10$ $(x-2)^2 + y^2 = 100$

Since $P(x, y)$ is on the parabola, $y^2 = 8x$. Substitute this into the equation: $(x-2)^2 + 8x = 100$ $x^2 - 4x + 4 + 8x = 100$ $x^2 + 4x - 96 = 0$ $(x+12)(x-8) = 0$

Possible values for $x$ are $-12$ and $8$. Since $y^2 = 8x$, $x$ must be non-negative ($x \ge 0$). Thus, we discard $x = -12$. So, $x = 8$.

Substitute $x = 8$ into $y^2 = 8x$: $y^2 = 8(8) = 64$ $y = \pm 8$

The coordinates of the points are $(8, 8)$ and $(8, -8)$.