Parabolas: Free Response

Write both distances as expressions in $x$ and $y$, then set them equal. The distance to the focus is $PF = \sqrt{(x-2)^2 + (y+4)^2}$, and the distance to the horizontal directrix $y=2$ is $|y-2|$, so

$\sqrt{(x-2)^2 + (y+4)^2} = |y - 2|.$

Both sides are nonnegative, so squaring is reversible:

$(x-2)^2 + (y+4)^2 = (y-2)^2.$

Expand the two squares in $y$: $(y+4)^2 = y^2 + 8y + 16$ and $(y-2)^2 = y^2 - 4y + 4$. Substituting,

$(x-2)^2 + y^2 + 8y + 16 = y^2 - 4y + 4.$

The $y^2$ terms cancel, and moving everything but $(x-2)^2$ to the right side gives

$(x-2)^2 = -4y + 4 - 8y - 16 = -12y - 12 = -12(y+1).$

So the equation is $(x-2)^2 = -12(y+1)$, already in the required form.

Match $(x-2)^2 = -12(y+1)$ against the template $(x-h)^2 = 4p(y-k)$. The vertex is $(h,k) = (2,-1)$, and the coefficient on the right is $4p$, so

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

Since $p<0$, the focus sits below the vertex and the curve opens downward.

Recover the focus and directrix from $h$, $k$, and $p$ and check them against the problem: the focus is $(h, k+p) = (2, -1-3) = (2,-4)$, and the directrix is $y = k - p = -1 - (-3) = 2$. Both agree exactly with the focus and directrix given at the start, confirming the derivation.

Squaring an equation can normally introduce extraneous solutions, because $A=B$ does not in general have the same solutions as $A^2=B^2$: for instance $A=3$ and $A=-3$ give the same square but are different equations. What rescues this particular squaring step is the nature of the two quantities being squared.

A square root, by its definition, is never negative, and an absolute value is never negative either. So both $\sqrt{(x-2)^2+(y+4)^2}$ and $|y-2|$ are nonnegative numbers for every $x$ and $y$. For two nonnegative real numbers $A$ and $B$, it is a genuine biconditional that

$A = B \;\Longleftrightarrow\; A^2 = B^2,$

because the function $t \mapsto t^2$ is one-to-one on the nonnegative numbers (it never sends two different nonnegative values to the same square). So no solution of the original equation is lost, and no new solution is gained: the squared equation has exactly the same solution set.

Match $x^2=16y$ against the template $x^2=4py$. The coefficient in front of $y$ is $4p$, not $p$ itself, so

$4p = 16 \;\Longrightarrow\; p = 4.$

With the vertex at the origin, the focus is $(0,p) = (0,4)$ and the directrix is $y=-p=-4$. The focal width is the absolute value of the same coefficient that was matched first, $|4p| = 16$.

The chord through the focus parallel to the directrix sits at the same height as the focus, $y=p=4$, and reaches $|2p|$ on either side of the axis of symmetry, so its endpoints are

$(-2p,\, p) = (-8,4), \qquad (2p,\, p) = (8,4).$

No substitution into $x^2=16y$ was needed to locate them. Checking one endpoint against the original equation,

$8^2 = 64 = 16(4),$

confirms it genuinely lies on the curve.

Substituting confirms the point is on the curve: $4^2 = 16 = 16(1)$.

By the distance formula, using the focus $(0,4)$,

$PF = \sqrt{(4-0)^2+(1-4)^2} = \sqrt{16+9} = \sqrt{25} = 5.$

By the shortcut,

$PF = y + p = 1 + 4 = 5.$

Both give $5$, and that is not a coincidence of this one point. The shortcut $PF=y+p$ is not a separate fact bolted onto the definition; for $p>0$ it is simply the distance from $(x,y)$ to the horizontal directrix $y=-p$, namely $|y-(-p)| = y+p$ for a point with $y \ge -p$. The defining condition of the parabola sets that distance equal to $PF$ for every point of the curve, so the two computations are two routes to the same number, not two independent checks.

The vertex reading is correct: matching $y+1$ and $x-3$ against $y-k$ and $x-h$ gives $(h,k)=(3,-1)$, exactly as the student found. The error is in the very next step.

The number $20$ sitting in the equation is $4p$, the focal width, not $p$ itself. Dividing correctly,

$4p = 20 \;\Longrightarrow\; p = 5.$

The focus is $p$ units from the vertex along the axis, so

$\text{focus } = (h+p, k) = (3+5,-1) = (8,-1),$

and the directrix is the same distance on the other side, $x = h-p = 3-5 = -2$. The student's answer of $(23,-1)$ came from adding the whole coefficient $20$ instead of its quarter.

A real number's square is never negative, so the left side of the equation satisfies

$(y+1)^2 \ge 0$

for every real $y$. The equation says this quantity equals $20(x-3)$, so

$20(x-3) \ge 0.$

Dividing by the positive number $20$ preserves the direction of the inequality:

$x - 3 \ge 0 \;\Longrightarrow\; x \ge 3.$

So every $(x,y)$ satisfying the equation has $x\ge3$, meaning the entire curve lies on or to the right of the vertical line $x=3$, which passes through the vertex. No point of the parabola can ever sit to the left of its own vertex.

The starting fact is the same as in part B: $(y-k)^2\ge0$ for every real $y$, and the equation says this equals $4p(x-h)$, so

$4p(x-h) \ge 0.$

This time the coefficient $4p$ is negative, and dividing an inequality by a negative number reverses its direction:

$x - h \le 0 \;\Longrightarrow\; x \le h.$

So every point of this parabola satisfies $x\le h$, meaning the whole curve lies on or to the left of the vertical line through the vertex, which is exactly what "opens left" means. The sign of $p$ was the only thing that changed between the two arguments, and it alone flipped both the inequality's direction and the curve's direction.

Factor $\tfrac14$ out of the two $x$-terms before completing the square:

$y = \tfrac14\bigl(x^2 - 4x\bigr) - 3 = \tfrac14\bigl((x-2)^2 - 4\bigr) - 3 = \tfrac14(x-2)^2 - 1 - 3 = \tfrac14(x-2)^2 - 4.$

The vertex form gives vertex $(2,-4)$. To isolate the squared term, add $4$ to both sides and multiply by the reciprocal of $\tfrac14$:

$y + 4 = \tfrac14(x-2)^2 \;\Longrightarrow\; (x-2)^2 = 4(y+4).$

From $(x-2)^2=4(y+4)$, the coefficient gives $4p=4$, so $p=1$. Since $p>0$, the parabola opens upward. The focus sits $p$ above the vertex, and the directrix the same distance below:

$\text{focus } (2,\,-4+1) = (2,-3), \qquad \text{directrix } y = -4-1 = -5.$

Rearranging vertex form $y=a(x-h)^2+k$ to isolate the square gives $(x-h)^2=\tfrac1a(y-k)$. Matching that against the standard conic form $(x-h)^2=4p(y-k)$ shows that the coefficient $\tfrac1a$ plays the role of $4p$, not $p$ on its own:

$4p = \tfrac1a.$

Solving for $p$ gives the correct relationship,

$p = \frac{1}{4a}.$

With $a=\tfrac14$, that is $p=\tfrac{1}{4(1/4)}=\tfrac11=1$, exactly the value found in part A. Treating $p$ as $\tfrac1a$ directly would have given $\tfrac1a=4$, four times too large, because it skips the division by $4$ that turning vertex form into the conic form requires.

Move everything except the $y$-terms to the other side:

$y^2 + 10y = 4x - 21.$

Half of $10$ is $5$, and $5^2=25$, so add $25$ to both sides to complete the square:

$y^2 + 10y + 25 = 4x - 21 + 25 \;\Longrightarrow\; (y+5)^2 = 4x + 4.$

Factoring the right side gives the required form, $(y+5)^2 = 4(x+1)$, with vertex $(-1,-5)$.

From $(y+5)^2=4(x+1)$, the coefficient gives $4p=4$, so $p=1$, and since $p>0$ the axis is horizontal and the curve opens to the right. The focus sits $p$ to the right of the vertex and the directrix the same distance to the left:

$\text{focus } (-1+1,\,-5) = (0,-5), \qquad \text{directrix } x = -1-1 = -2.$

Substitute $x=0$ into $(y+5)^2=4(x+1)$:

$(y+5)^2 = 4(0+1) = 4 \;\Longrightarrow\; y+5 = \pm2 \;\Longrightarrow\; y=-3 \text{ or } y=-7.$

Both $(0,-3)$ and $(0,-7)$ satisfy the equation, so the single vertical line $x=0$ meets the curve at two points. A graph is the graph of a function of $x$ exactly when every vertical line meets it at most once, and this line meets it twice, so the curve is not the graph of a function of $x$.

Solving the same equation for $x$ instead, dividing by $4$ and moving the constant,

$x = \tfrac14(y+5)^2 - 1.$

The right side is a single number for every real $y$, so this equation assigns exactly one $x$-value to each $y$-value: the curve is the graph of a function $x=g(y)$, even though it failed to be one of $y=f(x)$.