Parabolas: Additional Practice (Set 1)

This is a parabola with focus $(0,-3)$ and directrix $y = 3$. Write the distance condition for $(x,y)$ and square both sides, which is reversible since both sides are distances.

$\sqrt{x^2 + (y+3)^2} = |y - 3|$

$x^2 + y^2 + 6y + 9 = y^2 - 6y + 9$

The $y^2$ terms and the constants cancel, leaving $x^2 = -12y$.

The standard form confirms it: $4p = -12$ gives $p = -3$, so the focus is $3$ below the vertex at the origin, and the curve opens downward, away from the directrix above.

Complete the square, factoring the $3$ out of the $x$ terms.

$y = 3(x^2 - 4x) + 13 = 3\bigl((x-2)^2 - 4\bigr) + 13 = 3(x-2)^2 + 1$

The vertex is $(2,1)$. Isolate the square to expose $4p$: subtract $1$ and divide by $3$.

$(x-2)^2 = \tfrac13 (y - 1) \;\Longrightarrow\; 4p = \tfrac13 \;\Longrightarrow\; p = \tfrac{1}{12}$

The directrix is $p$ below the vertex, at $y = k - p$.

$y = 1 - \tfrac{1}{12} = \tfrac{11}{12}$

The distractor $\tfrac{13}{12}$ is the focus's height, on the other side of the vertex.

From $4p = 12$ we get $p = 3$, so this sideways parabola has focus $(3,0)$ and directrix $x = -3$. For a point on the curve, the distance to the focus equals the distance to the directrix, which for a vertical directrix is measured horizontally.

$PF = x + 3$

Set that equal to $7$:

$x + 3 = 7 \;\Longrightarrow\; x = 4$

Check it: at $x = 4$, $y^2 = 48$, and the distance from $(4, \sqrt{48})$ to $(3,0)$ is $\sqrt{1 + 48} = 7$.

Clear the fraction to reach the standard form. Multiplying by $-8$ gives $y^2 = -8x$.

$4p = -8 \;\Longrightarrow\; p = -2$

The square is on $y$, so the axis is horizontal and the focus is at $(p, 0)$, moving along the $x$-axis.

$(-2, \; 0)$

The negative $p$ opens the curve to the left, toward its focus. The distractor $(2,0)$ is where the directrix $x = 2$ crosses the axis, and $(0,-2)$ mistakenly treats the curve as vertical.

A horizontal axis with vertex $(1,-3)$ means $(y + 3)^2 = 4p(x - 1)$. Substitute the point $(9,1)$.

$(1 + 3)^2 = 4p(9 - 1) \;\Longrightarrow\; 16 = 32p \;\Longrightarrow\; p = \tfrac12$

For a horizontal axis the focus is at $(h + p, \; k)$, so it moves sideways from the vertex, never up.

$\left(1 + \tfrac12, \; -3\right) = \left(\tfrac32, -3\right)$

The answer $\left(1, -\tfrac52\right)$ makes the classic error of shifting the focus vertically away from the vertex, which is what the other standard form does.

Put the vertex at the origin with the dish opening upward, so its cross-section is $x^2 = 4py$. Across $60$ cm means the rim is $30$ cm either side of the axis, and $15$ cm deep puts the rim at height $15$, so $(30, 15)$ lies on the curve.

$30^2 = 4p(15) \;\Longrightarrow\; 900 = 60p \;\Longrightarrow\; p = 15$

The focus is $15$ cm from the vertex, which happens to be exactly level with the rim here. Using the full width $60$ in place of the half-width $30$ is the standard trap and would give a focus four times too far away.

Substitute the line into the parabola and collect everything on one side.

$x^2 = 4(x + k) \;\Longrightarrow\; x^2 - 4x - 4k = 0$

Tangency means the line meets the curve exactly once, so this quadratic must have a repeated root, which happens exactly when the discriminant is zero.

$(-4)^2 - 4(1)(-4k) = 16 + 16k = 0 \;\Longrightarrow\; k = -1$

Check it: with $k = -1$ the quadratic is $x^2 - 4x + 4 = (x-2)^2 = 0$, a double root at $x = 2$. For $k > -1$ the discriminant is positive and the line cuts the curve twice; for $k < -1$ it misses entirely.

Clear the fraction: multiplying by $6$ gives $x^2 = 6y$, which is $x^2 = 4py$ with

$4p = 6 \;\Longrightarrow\; p = \tfrac32.$

The focal width is $|4p| = 6$, the coefficient itself.

Confirm it by cutting the curve at the focus's height $y = \tfrac32$: then $x^2 = 6 \cdot \tfrac32 = 9$, so $x = \pm 3$ and the chord runs from $(-3, \tfrac32)$ to $(3, \tfrac32)$, a length of $6$. The distractor $\tfrac32$ is $p$, only a quarter of the width.

The focal width is $|4p|$, so start from the magnitude.

$|4p| = 10 \;\Longrightarrow\; |p| = \tfrac{10}{4} = \tfrac52$

Now fix the sign. Opening downward means the focus lies below the vertex, so $p$ is negative.

$p = -\tfrac52$

The curve is $x^2 = -10y$. The distractor $-10$ is $4p$ rather than $p$, and $\tfrac52$ has the right size but opens the curve the wrong way.

The tangent at $P$ bisects the segment from the focus $F$ to $D$, the foot of the perpendicular from $P$ to the directrix. Since $4p = 8$, we have $p = 2$, so the directrix is $y = -2$.

The foot of the perpendicular from $P = (4,2)$ to a horizontal line keeps the same $x$-coordinate.

$D = (4, -2)$

And the definition checks out: $PD = |2 - (-2)| = 4$, while $PF$, the distance from $(4,2)$ to $(0,2)$, is also $4$. Because $P$ is equidistant from $F$ and $D$, it lies on the perpendicular bisector of $FD$, which is the tangent, and that mirror is what reflects an incoming ray from $D$'s direction toward $F$.

An axis along the $y$-axis with the vertex at the origin means $x^2 = 4py$. Substitute the point $(3,-3)$.

$3^2 = 4p(-3) \;\Longrightarrow\; 9 = -12p \;\Longrightarrow\; p = -\tfrac34$

The directrix of $x^2 = 4py$ is the line $y = -p$.

$y = -\left(-\tfrac34\right) = \tfrac34$

The curve passes through a point below the vertex, so it opens downward, which puts the directrix above the vertex and the focus below it at $\left(0, -\tfrac34\right)$. The choice $y = -\tfrac34$ quotes that focus height instead of the directrix.

Find the curve first. Opening upward from the origin means $x^2 = 4py$, and the point $(6,3)$ is on it.

$6^2 = 4p(3) \;\Longrightarrow\; 36 = 12p \;\Longrightarrow\; p = 3$

So the parabola is $x^2 = 12y$ and its directrix is $y = -3$. For any point on the curve, the focal distance equals the distance to the directrix.

$PF = y + 3 = 12 \;\Longrightarrow\; y = 9$

The answer is $y = 9$, where $x^2 = 108$. Reading $12$ straight off as the height forgets that the focal distance is measured from $y = -3$, not from the vertex.