Exercises: Derive the Equation of a Parabola Given Focus and Directrix

$\sqrt{45}$, found by applying the distance formula from $(6, 2)$ to the point $(0, -3)$ where the line crosses the $y$-axis.

$5$, because the shortest path from a point to a horizontal line is straight down, a change of $2 - (-3)$ units in $y$ only.

$3$, because the line is $y = -3$ and the distance to a line named by a number is that number.

$\sqrt{85}$, found by applying the distance formula from $(6, 2)$ to the origin-side point $(-3, 0)$ on the line.

$y^2 + 16$, since squaring a sum squares each term.

$y^2 + 4y + 16$, using the first term, the coefficient $4$, and the square of $4$.

$y^2 + 8y + 16$, using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = y$ and $b = 4$.

$y^2 + 8y + 8$, doubling the $4$ for both the middle term and the constant.

$y = 4$, placing the directrix on the same side of the vertex as the focus.

$x = -4$, treating the parabola as opening sideways because the equation begins with $x^2$.

$y = -16$, reading the directrix straight off the coefficient $16$ without solving $4p = 16$.

$y = -4$, since $4p = 16$ gives $p = 4$ and the directrix of an upward parabola is $y = -p$.

Left, focus $(-3, 0)$ — the squared variable is $y$, so the axis is horizontal, and $4p = -12$ gives $p = -3$.

Right, focus $(3, 0)$ — the axis is horizontal, and the focus is taken $3$ units from the vertex on the positive side.

Down, focus $(0, -3)$ — the negative coefficient means the parabola opens downward from the origin.

Left, focus $(-12, 0)$ — the axis is horizontal and the coefficient $-12$ locates the focus directly.

$p = 2$, reading the focal length directly from the coefficient of $x^2$.

$p = 8$, computing $4a$ with $a = 2$ instead of dividing.

$p = \frac{1}{8}$, since $a = \frac{1}{4p}$ means $p = \frac{1}{4a} = \frac{1}{4(2)}$.

$p = \frac{1}{2}$, taking the reciprocal of the coefficient $a = 2$ without the factor of $4$.