Parabolas: Practice

Match against $(x - h)^2 = 4p(y - k)$. The vertex is $(3, -1)$ and the coefficient on the right is $4p$.

$4p = 4 \;\Longrightarrow\; p = 1$

The axis is vertical, so the focus is $p = 1$ unit above the vertex.

$(h, \; k + p) = (3, \; -1 + 1) = (3, 0)$

The distractor $(3,-2)$ is the directrix quoted as a point; the directrix is the line $y = -2$.

The square is on $y$, so the axis is horizontal and the form is $(y-k)^2 = 4p(x-h)$ with vertex $(1, -2)$.

$4p = -12 \;\Longrightarrow\; p = -3$

For a horizontal axis the directrix is the vertical line $x = h - p$.

$x = 1 - (-3) = 4$

Since $p < 0$ the curve opens left, so the directrix must be the line on the right, at $x = 4$.

The directrix is horizontal, so the axis is vertical and the form is $(x-h)^2 = 4p(y-k)$. The vertex is the midpoint between the focus and the directrix, measured along the axis $x = 2$.

$k = \frac{5 + 1}{2} = 3 \;\Longrightarrow\; \text{vertex } (2, 3)$

The signed distance from the vertex up to the focus is $p = 5 - 3 = 2$, so $4p = 8$.

$(x-2)^2 = 8(y-3)$

The focus is above the directrix, so $p > 0$ and the curve opens upward.

The directrix is vertical, so the axis is horizontal and the form is $(y-k)^2 = 4p(x-h)$. The vertex is midway between the focus and the directrix along the axis $y = 2$.

$h = \frac{-1 + 3}{2} = 1 \;\Longrightarrow\; \text{vertex } (1, 2)$

The signed distance from the vertex to the focus is $p = -1 - 1 = -2$, so $4p = -8$.

$(y-2)^2 = -8(x-1)$

The focus is to the left of the vertex, so the curve opens left, which is what the negative $p$ records.

Complete the square, factoring the leading coefficient out of the $x$ terms first.

$y = 2(x^2 - 4x) + 5 = 2\bigl((x-2)^2 - 4\bigr) + 5 = 2(x-2)^2 - 3$

In vertex form $y = a(x-h)^2 + k$ the vertex is $(h,k) = (2, -3)$. Multiplying the $-4$ back through by the $2$ is the step most often skipped: it turns $+5$ into $-3$.

These are the focus and directrix of a parabola, so test each candidate against the definition. For $(4,2)$, the distance to the point is

$\sqrt{(4-0)^2 + (2-2)^2} = \sqrt{16} = 4,$

and the distance to the line $y = -2$ is $|2 - (-2)| = 4$. The two agree, so $(4,2)$ is on the curve.

The others fail: $(2,4)$ is $\sqrt{8} \approx 2.83$ from the point but $6$ from the line, $(8,4)$ is $\sqrt{68} \approx 8.25$ from the point but $6$ from the line, and $(2,1)$ is $\sqrt{5} \approx 2.24$ from the point but $3$ from the line.

The focus is directly above the vertex, so the axis is vertical and the form is $(x-h)^2 = 4p(y-k)$ with $(h,k) = (-3,4)$. The signed distance from the vertex to the focus is

$p = 7 - 4 = 3 \;\Longrightarrow\; 4p = 12.$

$(x+3)^2 = 12(y-4)$

The coefficient is $4p = 12$, not $p = 3$, and the square sits on $x$ because the curve opens vertically.

The equation is quadratic in $y$ and linear in $x$, the signature of a sideways parabola. Isolate the $y$ terms and complete the square.

$y^2 - 4y = 8x - 28$

Half of $-4$ is $-2$, and $(-2)^2 = 4$, so add $4$ to both sides.

$(y - 2)^2 = 8x - 24 = 8(x - 3)$

Comparing with $(y-k)^2 = 4p(x-h)$ gives $h = 3$ and $k = 2$, so the vertex is $(3,2)$. Keep the order straight: $h$ is the $x$-coordinate even though the $y$ term was squared.

The square is on $y$, so the axis is horizontal. Matching $(y-k)^2 = 4p(x-h)$ gives vertex $(-2, 1)$ and

$4p = 4 \;\Longrightarrow\; p = 1.$

For a horizontal axis the focus is at $(h + p, \; k)$, moving along the axis, not up.

$(-2 + 1, \; 1) = (-1, 1)$

The distractor $(-3,1)$ is where the directrix crosses the axis, and $(-2,2)$ mistakenly moves the focus vertically.

Signals arriving from a distant satellite are effectively parallel to the axis of the dish. The reflective property says that every ray parallel to the axis, wherever it strikes the surface, reflects through one single point.

$\text{parallel ray} \;\xrightarrow{\;\text{reflects}\;}\; \text{focus}$

So all the collected energy is concentrated at the focus, which is where the receiver goes. That property is the reason the point is named the focus at all.

For a parabola with its vertex at the origin, the coefficient and the focus distance are linked by $a = \tfrac{1}{4p}$.

The focus is at $\left(0, \tfrac18\right)$, so $p = \tfrac18$.

$a = \frac{1}{4p} = \frac{1}{4 \cdot \tfrac18} = \frac{1}{\tfrac12} = 2$

Check it from the other side: $y = 2x^2$ rearranges to $x^2 = \tfrac12 y$, so $4p = \tfrac12$ and $p = \tfrac18$, as required. Reading $a = \tfrac18$ straight off the focus is the trap here.

The directrix is horizontal, so the axis is vertical and the directrix is the line $y = k - p$.

$-1 - p = 2 \;\Longrightarrow\; p = -3$

A negative $p$ says the focus lies below the vertex, which fits: the directrix is above it, and a parabola always runs away from its directrix.

$(h, \; k + p) = (2, \; -1 + (-3)) = (2, -4)$

The focus and the directrix sit on opposite sides of the vertex, each $3$ units away.