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Parabolas: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

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Problem 1 of 10

Problem 1 A point and line

The figure gives a focus and a directrix. Find the signed value of $p$ for their parabola.

The focus F and the directrix.

Text description of this figure

A grid with the x-axis from -5 to 3 and the y-axis from -2 to 6 on equal scales, a grid line and a label at every integer. A single point labeled F sits at x equals -1 and y equals 4. A dashed horizontal line, labeled directrix, is drawn along the x-axis, at height 0, across the whole grid. No parabola, vertex or distance is drawn.
1. Hint 1
  The vertex lies halfway from the focus to the directrix along the axis.
2. Hint 2
  Read the two heights, find their midpoint, and measure upward from that midpoint to the focus.
Answer

$p=2$ .

Full solution

The focus has height $4$ and the directrix has height $0$ , so the vertex height is
$k=\frac{4+0}{2}$ $k=2.$
The focus is above the vertex, so
$p=4-2$ $p=2.$
The vertex is also $2$ units above the directrix.

Answer

$p=2$ .

Key idea

The signed value of $p$ measures the displacement from the vertex toward the focus along the axis.
Problem 1: Mark complete
Problem 2 A measured chord

A parabola has a focal width of 14 cm. How far is its focus from its directrix?
1. Hint 1
  The focal width is four times the vertex-to-focus distance.
2. Hint 2
  The vertex lies halfway between the focus and directrix, so their separation is twice the vertex-to-focus distance.
Answer

7 cm.

Full solution

Write the focal width as $4|p|$ .
Then
$|p|=\frac{14}{4}$ $|p|=3.5.$
The focus and directrix lie on opposite sides of the vertex, so their separation is
$2|p|=7.$
The required distance is 7 cm.

Answer

7 cm.

Key idea

The separation of the focus and directrix is half the focal width.
Problem 2: Mark complete
Problem 3 Two quadratic graphs

The focus of $y=Ax^2$ , where $A>0$ , is twice as far from the origin as the focus of $y=\frac{x^2}{22}$ . Find $A$ .
1. Hint 1
  For an upward parabola, the focus distance is the reciprocal of four times the quadratic coefficient.
2. Hint 2
  How does the focus distance depend on the quadratic coefficient?
Answer

$A=\frac{1}{44}$ .

Full solution

For the given graph, $a=1/22$ , so
$p=\frac{1}{4(1/22)}$ $p=\frac{11}2.$
The new focus distance is $11$ .
Therefore
$A=\frac{1}{4(11)}$ $A=\frac{1}{44}.$
Its focus distance is $11$ , twice $11/2$ .

Answer

$A=\frac{1}{44}$ .

Key idea

For upward parabolas with the same vertex, the quadratic coefficient and focus distance vary reciprocally.
Problem 3: Mark complete
Problem 4 An unspecified line

The focus is $F=(7,9)$ , and $P=(10,13)$ lies on the parabola. The directrix is horizontal. Find every possible directrix.
1. Hint 1
  The distance from P to the focus fixes the required distance from P to the line.
2. Hint 2
  A horizontal line at height d is $|13-d|$ units from P.
Answer

$y=8$ or $y=18$ .

Full solution

The horizontal and vertical gaps from F to P are $3$ and $4$ , so
$PF=\sqrt{3^2+4^2}$ $PF=5.$
A horizontal directrix $y=d$ must satisfy
$|13-d|=5.$
Thus $d=8$ or $d=18$ .
Neither line contains F, which is at height $9$ , and each is exactly $5$ units from P, so both qualify.

Answer

$y=8$ or $y=18$ .

Key idea

A point-to-line distance can locate a horizontal line on either side of the point.
Problem 4: Mark complete
Problem 5 Moving a reflector

The parabola $y^2+10y+20x+65=0$ is translated so that its focus becomes $(-4,6)$ . Find the translation and the equation of the translated curve in standard form.
1. Hint 1
  A translation moves the focus and vertex by the same displacement.
2. Hint 2
  Complete the square in y to locate the original vertex and focus, then compare the focus coordinates.
Answer

3 units right and 11 units up; $(y-6)^2=-20(x-1)$ .

Full solution

Completing the square, $y^2+10y=(y+5)^2-25$ , so
$(y+5)^2=-20x-40$
that is,
$(y+5)^2=-20(x+2).$
The vertex is $(-2,-5)$ and $4p=-20$ , so $p=-5$ and the focus is $(-7,-5)$ .

Reaching $(-4,6)$ requires adding $(3,11)$ .
The vertex becomes $(1,6)$ and p is unchanged, giving
$(y-6)^2=-20(x-1).$
Its focus is $(1-5,6)$ , which is $(-4,6)$ , as required.

Answer

3 units right and 11 units up; $(y-6)^2=-20(x-1)$ .

Key idea

A translation moves the focus and vertex together while preserving the focal width.
Problem 5: Mark complete
Problem 6 A chord across the curve

A vertical parabola opens upward. Its chord through the focus, parallel to the directrix, has endpoints $(-5,7)$ and $(7,7)$ . Find its focus, directrix, and standard equation.
1. Hint 1
  The focus is the midpoint of this chord, and the chord length is $4p$ .
2. Hint 2
  Find p from the width, then move down by p from the focus to the vertex and by p again to the directrix.
Answer

Focus $(1,7)$ ; directrix $y=1$ ; $(x-1)^2=12(y-4)$ .

Full solution

The chord midpoint is $(1,7)$ and its length is $12$ , so
$p=\frac{12}{4}$ $p=3.$
The vertex is $(1,4)$ and the directrix is $y=1$ .
Hence
$(x-1)^2=12(y-4).$
At both endpoints the two sides equal $36$ , confirming the chord.

Answer

Focus $(1,7)$ ; directrix $y=1$ ; $(x-1)^2=12(y-4)$ .

Key idea

The chord through the focus parallel to the directrix is bisected by the focus and has length four times the vertex-to-focus distance.
Problem 6: Mark complete
Problem 7 A receiver adjustment

A receiver starts at the focus of $y=\frac{x^2}{18}$ . The replacement reflector has cross-section $(y+3)^2=30(x-8)$ , with coordinates in centimeters, and light travels leftward parallel to its axis. Give the receiver's horizontal and vertical movements to the new focus, and the new reflector's directrix.
1. Hint 1
  A reflector's focus is measured from its vertex along its axis.
2. Hint 2
  Find each vertex and signed p, then compare the focus coordinates; the directrix lies on the other side of the new vertex.
Answer

$\frac{31}{2}$ cm right and $\frac{15}{2}$ cm down; directrix $x=\frac12$ .

Full solution

The first curve is $x^2=18y$ , so $4p=18$ , $p=\frac92$ , and its focus is $(0,\frac92)$ .

The second has vertex $(8,-3)$ and $4p=30$ , so
$p=\frac{15}2.$
It opens to the right, so its focus is $(8+\frac{15}2,-3)$ , which is $(\frac{31}2,-3)$ , and its directrix is $x=8-\frac{15}2$ , that is, $x=\frac12$ .

From $(0,\frac92)$ to $(\frac{31}2,-3)$ the receiver moves $\frac{31}2$ cm right and $\frac{15}2$ cm down.
Rays entering from the open side parallel to the axis reflect through the focus, which is why it belongs there.

Answer

$\frac{31}{2}$ cm right and $\frac{15}{2}$ cm down; directrix $x=\frac12$ .

Key idea

A receiver collecting rays that enter a parabolic reflector's open side parallel to its axis belongs at its focus.
Problem 7: Mark complete
Problem 8 A proposed derivation

Let $t>0$ . A student starts with $\sqrt{x^2+(y+t)^2}=|y-t|$ and concludes $y=\frac{x^2}{4t}$ . Is the conclusion correct? Derive the correct equation in the form $y=ax^2$ .
1. Hint 1
  Both sides of the starting equation are distances, so squaring is reversible.
2. Hint 2
  After squaring, subtract $(y+t)^2$ from $(y-t)^2$ before dividing by a nonzero quantity.
Answer

No; $y=-\frac{x^2}{4t}$ , with $a=-\frac{1}{4t}$ .

Full solution

Squaring gives
$x^2+(y+t)^2=(y-t)^2.$
Isolate the squared x term:
$x^2=(y-t)^2-(y+t)^2.$
The constant and squared y terms cancel; the linear terms give
$x^2=-4ty.$
Since $t>0$ , divide by $-4t$ to obtain
$y=-\frac{x^2}{4t}.$
The focus $(0,-t)$ lies below the vertex, which checks the negative coefficient.

Answer

No; $y=-\frac{x^2}{4t}$ , with $a=-\frac{1}{4t}$ .

Key idea

Expanding the distance condition determines both the factor of four and the sign of a parabola's coefficient.
Problem 8: Mark complete
Problem 9 A reflected graph

The graph $y^2=-10x$ is reflected across the $y$ -axis. A student says the signed value of $p$ changes sign but the focal width stays the same. Decide whether this is correct and give both new values.
1. Hint 1
  Reflection across the y-axis replaces x by its opposite.
2. Hint 2
  Compare the new coefficient of x with $4p$ ; a width is a positive length.
Answer

Correct; $p=\frac52$ and focal width $10$ .

Full solution

Replacing x with $-x$ gives
$y^2=10x.$
Thus
$p=\frac{10}{4}$ $p=\frac52.$
Originally p was $-5/2$ .
The focal width is $|4p|=10$ before and after reflection, so the claim is correct.

Answer

Correct; $p=\frac52$ and focal width $10$ .

Key idea

Reflecting a horizontal parabola across the y-axis reverses its opening without changing its width.
Problem 9: Mark complete
Problem 10 A covered mirror

An upward parabolic mirror receives light traveling downward parallel to its axis. Its left half is covered, while its right half keeps the same shape. A student says the uncovered half will no longer direct all the rays it reflects through the original focus. Is this claim correct? Explain using the distance condition.
1. Hint 1
  Which geometric facts does the reflection argument use at a single point?
2. Hint 2
  For an uncovered point P, compare PF with the perpendicular distance from P to the directrix.
Answer

No; every ray reflected by the uncovered half still passes through the original focus.

Full solution

Covering part of the mirror does not move its focus, directrix, or remaining points.
At each uncovered point,
$PF=PD.$
The tangent there exchanges F and D as mirror images.
A ray arriving parallel to the axis travels along the line PD and reflects along PF toward F.
The covered half reduces the collected light but does not change where the remaining reflected rays meet.

Answer

No; every ray reflected by the uncovered half still passes through the original focus.

Key idea

The focusing property of a parabola holds at each reflecting point separately.
Problem 10: Mark complete

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