Parabolas: Additional Practice (Set 2)

Complete the square, factoring the $-2$ out of the $x$ terms.

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

The vertex is $(2,3)$. Now isolate the square to expose $4p$: subtract $3$ and divide by $-2$.

$(x-2)^2 = -\tfrac12 (y - 3) \;\Longrightarrow\; 4p = -\tfrac12 \;\Longrightarrow\; p = -\tfrac18$

The focus is $p$ from the vertex, and $p$ is negative because the curve opens downward.

$\left(2, \; 3 - \tfrac18\right) = \left(2, \tfrac{23}{8}\right)$

The distractor $\tfrac{25}{8}$ is where the directrix sits, on the other side of the vertex.

Put the vertex at the top of the arch, at $(0,5)$, so the curve is $x^2 = 4p(y - 5)$ and opens downward. The base is $20$ m wide, so the arch meets the ground $10$ m either side of the centre, at $(10, 0)$.

$10^2 = 4p(0 - 5) \;\Longrightarrow\; 100 = -20p \;\Longrightarrow\; 4p = -20$

The arch is therefore $x^2 = -20(y-5)$. Set $x = 5$:

$25 = -20(y - 5) \;\Longrightarrow\; y - 5 = -1.25 \;\Longrightarrow\; y = 3.75$

The height is $3.75$ m. Halving the distance from the centre does not halve the drop: the drop scales with the square of the horizontal distance, so it falls only a quarter as far as it does at the base.

Substitute the point to find $p$.

$(6-2)^2 = 4p(2+1) \;\Longrightarrow\; 16 = 12p \;\Longrightarrow\; p = \tfrac43$

The axis is horizontal with vertex $(-1, 2)$, so the directrix is the vertical line $x = h - p$.

$x = -1 - \tfrac43 = -\tfrac73$

Since $p > 0$ the curve opens right, so the directrix must lie to the left of the vertex, which confirms the negative value. The distractor $\tfrac13$ is the focus's $x$-coordinate, $h + p$.

The vertex is $(-2, 3)$, and for a vertical axis the directrix is the line $y = k - p$.

$3 - p = 5 \;\Longrightarrow\; p = -2$

The sign is the whole answer here. The directrix sits above the vertex, so the focus must sit below it, at $(-2, 1)$, and the curve opens downward. The distractor $-8$ is $4p$, the coefficient, not $p$.

From $4p = 8$ we get $p = 2$, so the focus is $(0, 2)$. The directrix is horizontal, so a chord parallel to it is the horizontal line through the focus, $y = 2$. Substitute that height into the equation.

$x^2 = 8(2) = 16 \;\Longrightarrow\; x = \pm 4$

The endpoints are $(-4, 2)$ and $(4, 2)$. Their separation is $8$, which is exactly the focal width $|4p| = 8$, as it must be.

Each has $|4p| = 4$, so $|p| = 1$: the focus is one unit from the vertex in every case. Find each vertex and step off $p$.

$x^2 = 4(y+1)\colon\; \text{vertex } (0,-1), \; p = 1 \;\Longrightarrow\; \text{focus } (0, -1 + 1) = (0,0)$

That is the one. The others land elsewhere: $x^2 = 4(y-1)$ has vertex $(0,1)$ and focus $(0,2)$; $x^2 = -4(y+1)$ has vertex $(0,-1)$ but $p = -1$, so its focus is $(0,-2)$; and $y^2 = 4(x-1)$ opens sideways from vertex $(1,0)$, giving focus $(2,0)$.

From $4p = 4$ we get $p = 1$, so the directrix is $y = -1$ and the focal distance of a point on the curve is $y + 1$.

$y + 1 = 5 \;\Longrightarrow\; y = 4$

Now find the matching $x$ values on the curve.

$x^2 = 4(4) = 16 \;\Longrightarrow\; x = \pm 4$

Both $(4,4)$ and $(-4,4)$ work, by symmetry about the axis. Check one: the distance from $(4,4)$ to the focus $(0,1)$ is $\sqrt{16 + 9} = 5$.

From $4p = 8$ we get $p = 2$, so the focus is $(0,2)$ and the directrix is $y = -2$. For a point on the curve the focal distance equals the distance to the directrix.

$PF = y + 2$

Every point of $x^2 = 8y$ has $y \ge 0$, so $y + 2$ is smallest when $y = 0$, at the vertex.

$PF_{\min} = 0 + 2 = 2$

The closest point is the vertex $(0,0)$, exactly $|p| = 2$ units from the focus. The point $(4,2)$ is on the curve and genuinely $4$ units from the focus, but it is not the closest; and $(0,2)$ is the focus itself, which does not lie on the parabola at all.

Find the original focus first. From $4p = 8$ we get $p = 2$, and the vertex is the origin, so the focus is $(0,2)$.

A translation slides every feature of the figure by the same vector, since it preserves all distances. Adding $3$ to $x$ and subtracting $2$ from $y$:

$(0 + 3, \; 2 - 2) = (3, 0)$

The image has equation $(x-3)^2 = 8(y+2)$, with vertex $(3,-2)$ and focus $2$ above it, at $(3,0)$, which agrees.

Solve for $y$ first, since $y$ appears to the first power.

$y = 2x^2 + 12x + 20$

Now complete the square, factoring the $2$ out of the $x$ terms.

$y = 2(x^2 + 6x) + 20 = 2\bigl((x+3)^2 - 9\bigr) + 20 = 2(x+3)^2 + 2$

In vertex form $y = a(x-h)^2 + k$, the vertex is $(-3, 2)$. Remember to multiply the $-9$ by the outside $2$ before combining it with the $20$, which is what turns $20$ into $2$.

Opening right from the origin means $y^2 = 4px$ with $p > 0$. That chord is the focal chord, so its endpoints lie at the focus's $x$-coordinate, giving $p = 1$ straight away.

Confirm with the curve itself. At $x = p$ the equation gives

$y^2 = 4p \cdot p = 4p^2 \;\Longrightarrow\; y = \pm 2p,$

so the endpoints are $(p, \pm 2p)$. Matching $(1, \pm 2)$ gives $p = 1$ from both coordinates, a consistent answer.

$y^2 = 4(1)x = 4x$

The chord's length is $4 = |4p|$, the focal width, as expected.

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

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

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

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

Verify from the definition: $\sqrt{(x-2)^2 + (y-1)^2} = |y+3|$ squares to $(x-2)^2 + y^2 - 2y + 1 = y^2 + 6y + 9$, which simplifies to $(x-2)^2 = 8y + 8$. The two agree.