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Ellipses: 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 The oval on the grid

Find the distance between the foci of the ellipse in the figure.

The ellipse.

Text description of this figure

A grid with both axes from -7 to 7 on equal scales, a grid line and a label at every integer. A single ellipse, taller than it is wide, is centered where the axes cross: it meets the x-axis at x equals -5 and x equals 5, and the y-axis at y equals -6 and y equals 6. No equation, vertex, focus or center is marked.
1. Hint 1
  Read the two semiaxis lengths from the graph.
2. Hint 2
  The square of the center-to-focus distance is the larger squared semiaxis minus the smaller one.
Answer

$2\sqrt{11}$ units.

Full solution

The ellipse reaches $6$ units above and below its center and $5$ units to either side.
Thus $a=6$ and $b=5$ , so
$c^2=36-25$ $c^2=11.$
The center-to-focus distance is $c=\sqrt{11}$ , and the foci are on opposite sides of the center, giving
$2c=2\sqrt{11}.$

Answer

$2\sqrt{11}$ units.

Key idea

The two foci are separated by twice the distance found from the difference of the squared semiaxes.
Problem 1: Mark complete
Problem 2 A relocated outline

The ellipse $\frac{x^2}{20}+\frac{y^2}{8}=1$ is translated 2 units left and 3 units up. Write the new equation in standard form.
1. Hint 1
  A translation changes the center but preserves both semiaxis lengths.
2. Hint 2
  A center coordinate h appears as $x-h$ , and a center coordinate k appears as $y-k$ .
Answer

$\frac{(x+2)^2}{20}+\frac{(y-3)^2}{8}=1$ .

Full solution

The old center $(0,0)$ becomes $(-2,3)$ .
Replace x by $x+2$ and y by $y-3$ , giving
$\frac{(x+2)^2}{20}+\frac{(y-3)^2}{8}=1.$
The denominators stay $20$ and $8$ , so the lengths are unchanged.

Answer

$\frac{(x+2)^2}{20}+\frac{(y-3)^2}{8}=1$ .

Key idea

Translating an ellipse changes its center expressions while leaving its semiaxis lengths fixed.
Problem 2: Mark complete
Problem 3 Two distance measurements

The major axis of an ellipse has length 20 units. At a point on the ellipse, the distances to the foci are in the ratio $2:3$ . Find those two distances.
1. Hint 1
  The sum of the focal distances equals the full major-axis length.
2. Hint 2
  Write the two distances as $2u$ and $3u$ , with $u>0$ .
Answer

8 units and 12 units.

Full solution

The constant sum is $20$ .
Writing the distances as $2u$ and $3u$ gives
$5u=20.$
Thus $u=4$ , so the distances are $8$ and $12$ units.
They sum to $20$ and have the required ratio.

Answer

8 units and 12 units.

Key idea

A focal-distance ratio and the constant focal sum determine the two distances.
Problem 3: Mark complete
Problem 4 An equation from distances

For $P=(x,y)$ , let $r=\sqrt{x^2+(y-2)^2}$ and $s=\sqrt{x^2+(y+2)^2}$ . Starting from $r+s=12$ , derive the ellipse equation in standard form by squaring to remove the radicals. Explain why the resulting equation adds no extra points.
1. Hint 1
  Isolate one distance before the first squaring; the difference of their squared expressions is simple.
2. Hint 2
  The first squaring can be rearranged into $r=6-y/3$ . Square this, then collect the squared terms.
3. Hint 3
  On the resulting ellipse, check the range of y so that both recovered distances stay positive.
Answer

$\frac{x^2}{32}+\frac{y^2}{36}=1$ ; no extra points.

Full solution

Isolate $s=12-r$ and square:
$s^2=144-24r+r^2.$
The distance expressions give
$s^2-r^2=8y.$
Therefore
$24r=144-8y,$
so $r=6-y/3$ .

Square again and substitute the expression for $r^2$ :
$x^2+(y-2)^2=(6-y/3)^2.$
Expansion cancels the linear y terms, leaving
$x^2+\frac89y^2=32.$
Dividing by the positive number $32$ gives
$\frac{x^2}{32}+\frac{y^2}{36}=1.$

Conversely, the final equation forces $|y|\le6$ .
On it, $r^2=(6-y/3)^2$ and $s^2=(6+y/3)^2$ , and both $6-y/3$ and $6+y/3$ are at least $4$ , so $r=6-y/3$ and $s=6+y/3$ .
Hence $r+s=12$ , and no extra points were introduced.

Answer

$\frac{x^2}{32}+\frac{y^2}{36}=1$ ; no extra points.

Key idea

Checking the signs of recovered distances completes an ellipse derivation that squares twice.
Problem 4: Mark complete
Problem 5 A triangle inside the oval

In the figure, C is an ellipse's center, F is one focus, and B is a co-vertex. Find the major-axis length and explain how the triangle CFB determines it.

The center C, a focus F and a co-vertex B of an ellipse.

Text description of this figure

A grid with the x-axis from -8 to 1 and the y-axis from -7 to 0 on equal scales, a grid line and a label at every integer. Three labeled points form a right triangle: C at x equals -6 and y equals -5, F at x equals -1 and y equals -5 on the same horizontal line, and B at x equals -6 and y equals -3, directly above C. The segments CF, CB and FB are drawn, and a small square marks the right angle at C. The ellipse itself is not drawn, and no length or coordinate is written.
1. Hint 1
  A co-vertex is equally far from both foci, so each focal distance there is a.
2. Hint 2
  Read the two perpendicular legs from the grid and find the hypotenuse.
Answer

$2\sqrt{29}$ units.

Full solution

The center is $(-6,-5)$ , F is $(-1,-5)$ , and B is $(-6,-3)$ .
Thus CF is $5$ units and CB is $2$ units.
They are perpendicular, so
$BF=\sqrt{5^2+2^2}$ $BF=\sqrt{29}.$
At a co-vertex each focal distance equals $a$ , so $a=\sqrt{29}$ and the full major axis has length $2a=2\sqrt{29}$ units.

Answer

$2\sqrt{29}$ units.

Key idea

For an ellipse with distinct foci, the center, a focus and a co-vertex form a right triangle whose hypotenuse is the semimajor axis.
Problem 5: Mark complete
Problem 6 A changed tracing length

Two pins stay at $(-6,1)$ and $(-6,9)$ . A taut string attached to the pins initially has length 18 units and is replaced by a string of length 22 units. Find the standard equation of the new ellipse and the increase in its minor-axis length.
1. Hint 1
  The pins keep c fixed, while each string length determines a.
2. Hint 2
  Compute the old and new values of b using $b^2=a^2-c^2$ , and compare the full minor-axis lengths.
Answer

New ellipse: $\frac{(x+6)^2}{105}+\frac{(y-5)^2}{121}=1$ ; minor-axis increase $2(\sqrt{105}-\sqrt{65})$ units.

Full solution

The pins give center $(-6,5)$ and a vertical major axis, with focal distance $c=4$ .
Initially $a=9$ , so $b^2=81-16$ , which is $65$ .
With the new string, $a=11$ , so
$b^2=121-16$ $b^2=105.$
The new equation is
$\frac{(x+6)^2}{105}+\frac{(y-5)^2}{121}=1.$
The minor axis grows from $2\sqrt{65}$ to $2\sqrt{105}$ , giving the stated increase.

Answer

New ellipse: $\frac{(x+6)^2}{105}+\frac{(y-5)^2}{121}=1$ ; minor-axis increase $2(\sqrt{105}-\sqrt{65})$ units.

Key idea

Increasing the focal sum with fixed foci changes both semiaxes while keeping the focal separation fixed.
Problem 6: Mark complete
Problem 7 A family of outlines

For $k>0$ , the ellipse $\frac{x^2}{k+15}+\frac{y^2}{k}=1$ must have eccentricity $\frac14$ . Find $k$ and the coordinates of its foci.
1. Hint 1
  The difference of the two denominators fixes the focal distance for every value of k.
2. Hint 2
  Set the ratio of c to a equal to the required eccentricity.
Answer

$k=225$ ; foci $(-\sqrt{15},0)$ and $(\sqrt{15},0)$ .

Full solution

The larger denominator is $k+15$ , so the major axis is horizontal.
The focal distance satisfies
$c^2=(k+15)-k$ $c^2=15.$
Thus $c=\sqrt{15}$ , and $e=c/a=\frac14$ gives $a=4\sqrt{15}$ , so $a^2=240$ .
Then $k+15=240$ and $k=225$ .
The foci are $(\pm\sqrt{15},0)$ , and $\sqrt{15}/\sqrt{240}=\frac14$ checks the required ratio.

Answer

$k=225$ ; foci $(-\sqrt{15},0)$ and $(\sqrt{15},0)$ .

Key idea

The difference of an ellipse's squared semiaxes fixes c even when both semiaxes change.
Problem 7: Mark complete
Problem 8 A negative right side

A student says $-3(x+5)^2-7(y+6)^2=-84$ has no real points because the right side is negative. Is that correct? Give the major-axis direction and length if an ellipse exists.
1. Hint 1
  Compare the signs of the two sides before judging that no point exists.
2. Hint 2
  Divide every term by the same nonzero constant to reach standard form.
Answer

No; it is an ellipse with a horizontal major axis of length $4\sqrt7$ units.

Full solution

Both sides are negative, so dividing every term by $-84$ gives
$\frac{(x+5)^2}{28}+\frac{(y+6)^2}{12}=1.$
The larger denominator, $28$ , is under the x term, so the major axis is horizontal, $a=\sqrt{28}=2\sqrt7$ , and its length is $2a=4\sqrt7$ units.

Answer

No; it is an ellipse with a horizontal major axis of length $4\sqrt7$ units.

Key idea

Dividing both sides by the same negative number keeps an equation's solutions, so a negative right side alone does not rule out an ellipse.
Problem 8: Mark complete
Problem 9 Sliding the foci

An ellipse and both of its foci are translated by the same displacement. A student claims that its eccentricity stays unchanged even though its center moves. Is the claim correct? Justify it from the lengths in the eccentricity ratio.
1. Hint 1
  A translation preserves distances between points.
2. Hint 2
  Consider the center-to-focus distance and half the major-axis length separately.
Answer

Yes; the eccentricity stays unchanged.

Full solution

A translation preserves the center-to-focus distance c and the semimajor-axis length a.
The new lengths therefore satisfy
$\frac{c_{\mathrm{new}}}{a_{\mathrm{new}}}=\frac ca.$
This is the eccentricity before and after the move, so the claim is correct.
Changing the center alone changes position, not this shape ratio.

Answer

Yes; the eccentricity stays unchanged.

Key idea

A translation preserves eccentricity because it preserves both lengths in c divided by a.
Problem 9: Mark complete
Problem 10 Pins brought together

Two foci move toward each other until they coincide at $(-2,4)$ . Throughout the motion, points on the traced curve have focal distances whose sum is 18 units. Describe the final curve, give its equation, and explain its eccentricity.
1. Hint 1
  When the foci coincide, the two focal distances from any point are equal.
2. Hint 2
  The sum of two equal distances fixes one distance from the shared focus.
Answer

Circle centered at $(-2,4)$ with radius 9 units; $(x+2)^2+(y-4)^2=81$ ; $e=0$ .

Full solution

At the final position, each of the two focal distances equals the distance d from $(-2,4)$ .
Thus
$2d=18,$
so $d=9$ .
The locus is the circle
$(x+2)^2+(y-4)^2=81.$
The focal separation is zero, so $c=0$ while $a=9$ , giving $e=0$ .

Answer

Circle centered at $(-2,4)$ with radius 9 units; $(x+2)^2+(y-4)^2=81$ ; $e=0$ .

Key idea

Coincident foci and a fixed positive focal sum define a circle with zero eccentricity.
Problem 10: Mark complete

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