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Precalculus · Mathematics
Conics · Conics

Ellipses

Before you start, make sure you know: Loci (Geometry)
Before you start, make sure you know: Pythagorean Theorem (Geometry)

Push two thumbtacks into a board, loop a piece of string around them, pull the string tight with a pencil, and slide the pencil around. You have just drawn an ellipse. The same curve is the path of every planet around the Sun, with the Sun at one focus rather than at the center. It is also the shape of rooms where a whisper at one spot can be heard clearly at another. Why does the pencil trace such a perfect oval, and why do those two special spots work? Let's find out.

The ellipse as a locus

The string stays tight, so the pencil is always at a point $P$ where (distance to thumbtack 1) + (distance to thumbtack 2) equals the string length minus the gap between the tacks. That total never changes. This is the definition.

An ellipse with foci F1 and F2 and a point P on it; the segments from P to the two foci have lengths d1 and d2, and their sum is always 2a. F₁ F₂ P d₁ d₂ d₁ + d₂ = 2a for every point P

The constant $2a$ equals the string length minus the gap between the tacks (the distance $2c$ between the foci). It is also the length of the major axis, the longest chord of the ellipse.

Now let's turn the definition into an equation. This is exactly what "find the locus" problems ask you to do.

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Find the equation of the locus of the points in the plane $P(x, y)$ whose sum of distances to the points $F_1(-3, 0)$ and $F_2(3, 0)$ is always equal to $10$. Simplify…

Find the equation of the locus of the points in the plane $P(x, y)$ whose sum of distances to the points $F_1(-3, 0)$ and $F_2(3, 0)$ is always equal to $10$. Simplify the expression until you obtain the standard equation of an ellipse.

Solve

Standard equation centered at the origin

When the center is the origin and the axes of the ellipse lie along the coordinate axes, the equation is very clean.

An ellipse centered at the origin with semi-major axis a, semi-minor axis b, and foci F1 and F2 at distance c from the center; the right triangle with legs b and c and hypotenuse a shows that a squared equals b squared plus c squared. F₁ F₂ b c a foci (±c, 0) · a² = b² + c²

Some books call all four endpoints of the axes "vertices." Here the vertices are the endpoints of the major axis, and the co-vertices are the endpoints of the minor axis. Some problems use the phrase focal distance for the distance $2c$ between the two foci.

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Given the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$, determine the coordinates of its vertices, those of its foci, and the value of its eccentricity.

Given the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$, determine the coordinates of its vertices, those of its foci, and the value of its eccentricity.

Solve
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Write the standard equation of an ellipse centered at the origin of coordinates, knowing that one of its foci is the point $F(4, 0)$ and that the length of its major…

Write the standard equation of an ellipse centered at the origin of coordinates, knowing that one of its foci is the point $F(4, 0)$ and that the length of its major axis is $10$.

Solve

Eccentricity: how flat is the ellipse?

Two ellipses can both have $a = 5$ and still look very different. The foci can be close together (nearly a circle) or far apart (long and thin). Eccentricity measures this.

Three ellipses with different eccentricity: nearly circular with e about 0.2, medium with e about 0.7, and very long and thin with e about 0.97. e ≈ 0.2 (nearly a circle) e ≈ 0.7 e ≈ 0.97 (long and thin)

Try it yourself. Here $a = 5$ is fixed. Drag $b$ from $5$ down to $0.5$ and watch the shape, then think about where the foci would be.

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Calculate the eccentricity of the ellipse whose equation is $16x^2 + 25y^2 = 400$.

Calculate the eccentricity of the ellipse whose equation is $16x^2 + 25y^2 = 400$.

Solve
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Find the equation of an ellipse centered at the origin, whose foci are on the y-axis, knowing that the focal distance is $8$ units and the eccentricity is $0,8$.

Find the equation of an ellipse centered at the origin, whose foci are on the y-axis, knowing that the focal distance is $8$ units and the eccentricity is $0,8$.

Solve
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The eccentricity of an ellipse is $\frac{1}{3}$ and its semi-minor axis measures $4\sqrt{2}$. Calculate the measure of its semi-major axis.

The eccentricity of an ellipse is $\frac{1}{3}$ and its semi-minor axis measures $4\sqrt{2}$. Calculate the measure of its semi-major axis.

Solve

Ellipses not centered at the origin

If the center is $(h, k)$, replace $x$ with $x - h$ and $y$ with $y - k$. This is the same shift you used for circles.

An ellipse whose center is the point (h, k), away from the origin; dashed segments show the horizontal distance h and the vertical distance k from the axes to the center, and the two foci lie on the horizontal major axis. (h, k) h k x y

Problems often hand you a general equation $Ax^2 + By^2 + Cx + Dy + E = 0$. To get the center and axes, complete the square in $x$ and in $y$ separately.

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An ellipse has its center at the point $C(-2, 3)$, its major axis is parallel to the x-axis and measures $12$ units, and its minor axis measures $8$ units. Write its…

An ellipse has its center at the point $C(-2, 3)$, its major axis is parallel to the x-axis and measures $12$ units, and its minor axis measures $8$ units. Write its equation in standard form.

Solve
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From the equation of the ellipse $\frac{(x-1)^2}{16} + \frac{(y+4)^2}{9} = 1$, state the coordinates of its center and the length of its semi-axes.

From the equation of the ellipse $\frac{(x-1)^2}{16} + \frac{(y+4)^2}{9} = 1$, state the coordinates of its center and the length of its semi-axes.

Solve
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Given the general equation $9x^2 + 4y^2 - 36x + 24y + 36 = 0$, prove that it corresponds to an ellipse. To do this, express it in standard form and determine the…

Given the general equation $9x^2 + 4y^2 - 36x + 24y + 36 = 0$, prove that it corresponds to an ellipse. To do this, express it in standard form and determine the coordinates of its center, its vertices, and its foci.

Solve

Points and lines versus an ellipse

Is a point inside, on, or outside?

The ellipse is the set where $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Moving a point outward makes $x$ and $y$ larger in absolute value, so the left side grows. That gives a quick test.

The ellipse x squared over 36 plus y squared over 16 equals 1 with three points: A inside, B on the ellipse, and C outside. A B C A inside · B on the ellipse · C outside

Where does a line cross the ellipse?

Substitute the line into the ellipse equation. You get a quadratic, and its discriminant tells you what happens.

The same idea works when a condition ties $x$ and $y$ together and you need a point on the ellipse. Substitute the condition into the equation, solve, then go back for the other coordinate. For example, on $x^2 + 4y^2 = 20$ with $y = 3x$ you get $x^2 + 36x^2 = 20$, which gives $x^2 = \tfrac{20}{37}$.

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Check whether the points $P(2, -3)$, $Q(5, 0)$, and $R(4, 2)$ are interior, exterior, or belong to the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Check whether the points $P(2, -3)$, $Q(5, 0)$, and $R(4, 2)$ are interior, exterior, or belong to the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Solve
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Study the relative position of the line with equation $2x + y - 5 = 0$ and the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. If they intersect, find the coordinates of…

Study the relative position of the line with equation $2x + y - 5 = 0$ and the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. If they intersect, find the coordinates of the points of intersection.

Solve
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In a clockwork mechanism, a gear has an elliptical shape. It is known that its boundary equation is given by $x^2 + 4y^2 = 16$. It is desired to place a small pin at a…

In a clockwork mechanism, a gear has an elliptical shape. It is known that its boundary equation is given by $x^2 + 4y^2 = 16$. It is desired to place a small pin at a point on the ellipse whose x-coordinate is twice its y-coordinate. Calculate the coordinates of this point (assuming both are positive) and the exact distance from this point to the center of the gear.

Solve

Ellipses in the real world

To use an ellipse in a real problem: place the center at the origin, line up the major axis with an axis, find $a$ and $b$, and write the equation. Then answer the question with the equation.

Orbits: closest and farthest

If the Sun is at one focus, the point of the orbit closest to the Sun is a vertex, and so is the farthest point. The distances are:

$$\text{closest} = a - c, \qquad \text{farthest} = a + c.$$

Adding gives $2a$, and subtracting gives $2c$.

Arches

A semi-elliptical arch sitting on a horizontal base, with the semi-major axis a along the base and the height b measured at the center. b a Semi-ellipse: width 2a, height b

Whispering galleries

All the paths from one focus to the wall and then to the other focus have the same length, $2a$. Sound that leaves one focus in any direction therefore arrives at the other focus at the same instant, and the waves reinforce each other. That is why a whisper at one focus can be heard at the other.

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Mercury's orbit around the Sun is an ellipse in which the Sun occupies one of its foci. The minimum distance from Mercury to the Sun (perihelion) is $46$ million miles,…

Mercury's orbit around the Sun is an ellipse in which the Sun occupies one of its foci. The minimum distance from Mercury to the Sun (perihelion) is $46$ million miles, and the maximum distance (aphelion) is $70$ million miles. Calculate the eccentricity of Mercury's orbit and the length of its semi-minor axis in millions of miles.

Sun Perihelion Aphelion 46 Mmi 70 Mmi
Solve
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A stone bridge over a river has the shape of a semi-elliptical arch. The width of the river (major axis of the ellipse) is $20\text{ m}$ and the maximum height of the…

A stone bridge over a river has the shape of a semi-elliptical arch. The width of the river (major axis of the ellipse) is $20\text{ m}$ and the maximum height of the arch at the center of the river is $6\text{ m}$. Calculate at what height the arch is at a distance of $4\text{ m}$ from the center of the river.

20 ft 6 ft 4 ft h = ?
Solve
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In a whispering gallery, which has the shape of an ellipse, two people located at the foci can hear each other perfectly even if they speak in a whisper. If the gallery…

In a whispering gallery, which has the shape of an ellipse, two people located at the foci can hear each other perfectly even if they speak in a whisper. If the gallery has a total length of $34\text{ m}$ and a maximum width of $16\text{ m}$, at what distance from the center of the room must each person be located to take advantage of this acoustic phenomenon?

Solve

Distances from a point to the foci

For any point $P$ on the ellipse, $d_1 + d_2 = 2a$, where $d_1$ and $d_2$ are its distances to the two foci. This one fact solves many problems. If you also know an angle, the Law of Cosines gives a second equation.

In the triangle $F_1 P F_2$ the sides are $d_1$, $d_2$, and $2c$ (the distance between the foci). If $\theta$ is the angle at $F_1$, the Law of Cosines says $d_2^2 = d_1^2 + (2c)^2 - 2 d_1 (2c)\cos\theta$. Replace $d_2$ with $2a - d_1$ and the $d_1^2$ terms cancel, leaving a linear equation in $d_1$.

The ellipse x squared over 25 plus y squared over 16 equals 1 with foci F1 and F2 and a point P on the upper half; the triangle F1 P F2 has a 60 degree angle at F1 and sides d1, d2 and the focal distance 2c. F₁ F₂ P d₁ d₂ 60° 2c
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A main square has the shape of an ellipse. A straight path is to be paved joining the leftmost focus, $F_1$, with a point $P$ on the ellipse located in the first…

A main square has the shape of an ellipse. A straight path is to be paved joining the leftmost focus, $F_1$, with a point $P$ on the ellipse located in the first quadrant. It is known that the major axis of the square measures $100\text{ m}$ and the minor axis $60\text{ m}$. If the angle formed by the line segment $F_1P$ with the positive x-axis is $60^\circ$, calculate the exact length of the paved path.

F₁ F₂ P d = ? 60° 2b = 60 ft 2a = 100 ft
Solve

Building an equation from points and conditions

An ellipse through two given points

If the ellipse is centered at the origin with axes along the coordinate axes, write it as $\dfrac{x^2}{p} + \dfrac{y^2}{q} = 1$, or equivalently $Ax^2 + By^2 = 1$ with $A = \dfrac{1}{p}$ and $B = \dfrac{1}{q}$. Each point gives a linear equation in $A$ and $B$, so two points give a system you can solve by substitution or elimination.

Rectangles inscribed in an ellipse

A rectangle with sides parallel to the axes and its corners on the ellipse is symmetric about both axes. If one corner is $(x, y)$ with $x, y > 0$, the other corners are $(\pm x, \pm y)$. So the rectangle is $2x$ wide and $2y$ tall, and its area is $4xy$.

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An ellipse is centered at the origin and its axes of symmetry coincide with the Cartesian coordinate axes. If it is known that the ellipse passes through the points…

An ellipse is centered at the origin and its axes of symmetry coincide with the Cartesian coordinate axes. If it is known that the ellipse passes through the points $A(2, \sqrt{3})$ and $B(\sqrt{2}, \frac{\sqrt{15}}{2})$, find the equation of the ellipse and calculate the exact distance between its foci.

Solve
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A rectangular window is to be designed to be inscribed in an elliptical opening with equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$. The sides of the rectangle must be…

A rectangular window is to be designed to be inscribed in an elliptical opening with equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$. The sides of the rectangle must be parallel to the axes of the ellipse. If the rectangle is to have a total area of $24\text{ m}^2$, what must its exact dimensions be?

Solve

Tangent lines

A tangent line touches the ellipse at exactly one point. From the line-and-ellipse test, that means the discriminant is $0$. Doing the algebra once in general gives a handy shortcut.

This formula does not cover vertical lines, which have no slope. For this ellipse the vertical tangents are $x = \pm\sqrt{p}$.

The ellipse x squared over 16 plus y squared over 4 equals 1 with two tangent lines drawn from the external point P at (0, 4), touching the ellipse at two symmetric points. P(0, 4) Two tangent lines from an outside point

If a problem asks for a parameter, like "for which $k$ is $x + 2y + k = 0$ tangent?", rewrite the line as $y = mx + (\text{intercept})$ first, and then use the condition on the intercept.

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Find the equations of the tangent lines to the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$ that pass through the exterior point $P(0, 5)$.

Find the equations of the tangent lines to the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$ that pass through the exterior point $P(0, 5)$.

Solve
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Determine the equations of the tangent lines to the ellipse $x^2 + 2y^2 = 18$ that are parallel to the line with equation $x - y + 4 = 0$.

Determine the equations of the tangent lines to the ellipse $x^2 + 2y^2 = 18$ that are parallel to the line with equation $x - y + 4 = 0$.

Solve
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Calculate the values of the parameter $k$ so that the line $x + 2y + k = 0$ is tangent to the ellipse with equation $x^2 + 4y^2 = 20$.

Calculate the values of the parameter $k$ so that the line $x + 2y + k = 0$ is tangent to the ellipse with equation $x^2 + 4y^2 = 20$.

Solve

Focus and directrix

There is a second way to define an ellipse that does not use two foci at once.

The ellipse x squared over 16 plus y squared over 12 equals 1 with the focus F at (2, 0) and the vertical directrix x equals 8; a point P on the ellipse is 4 units from the focus and 8 units from the directrix. F P 4 8 directrix x = 8
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Determine the locus of the points in the plane whose distance to the point $F(4, 0)$ is equal to half their distance to the line with equation $x - 16 = 0$. Identify the…

Determine the locus of the points in the plane whose distance to the point $F(4, 0)$ is equal to half their distance to the line with equation $x - 16 = 0$. Identify the resulting conic and calculate its center and the length of its semi-axes.

Solve

More loci that give ellipses

Stretching or squashing a circle

If you keep every $x$-coordinate of a circle and multiply every $y$-coordinate by the same number $k$ with $0 < k < 1$, you squash the circle into an ellipse. Substitute the old coordinate (written in terms of the new one) into the circle's equation.

Circles tangent to a given circle

Suppose a circle has center $P$ and radius $r$, passes through a fixed point $A$, and is internally tangent to a big circle with center $O$ and radius $R$. Then $d(P, A) = r$, because $A$ is on the moving circle. Also $d(O, P) = R - r$, because the circles touch from the inside. Add them to eliminate $r$.

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Consider the circle with equation $x^2 + y^2 = 100$. To each point $A(x, y)$ on the circle is associated a point $P(x, y')$ that has the same x-coordinate, but its…

Consider the circle with equation $x^2 + y^2 = 100$. To each point $A(x, y)$ on the circle is associated a point $P(x, y')$ that has the same x-coordinate, but its y-coordinate is $60\%$ of the y-coordinate of $A$. Prove that the locus of the points $P$ is an ellipse and determine its equation, its center, and its foci.

Solve
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Find the locus of the centers of the circles that are internally tangent to the circle with equation $x^2 + y^2 = 100$ and that pass through the fixed point $A(6, 0)$.…

Find the locus of the centers of the circles that are internally tangent to the circle with equation $x^2 + y^2 = 100$ and that pass through the fixed point $A(6, 0)$. Prove that this locus is an ellipse, find its standard equation, and determine the coordinates of its foci.

Solve

Back to the beginning: the string has fixed length, so the two pieces of string from the pencil to the tacks always add up to the same number. That is exactly the definition $d_1 + d_2 = 2a$. In a whispering gallery, every sound path from one focus to the wall and on to the other focus also has length $2a$. All the waves arrive together, and the whisper carries.

Summary

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