The Ellipse

Before you start, make sure you know: Geometric loci (Secondary Year 3)

Before you start, make sure you know: The Pythagorean theorem (Secondary Year 3)

Stick two thumbtacks into a board, tie the ends of a string to each one, and draw with a pencil while keeping the string taut the whole time: you are drawing an ellipse. It is the curve of the planets' orbits (with the Sun at a focus, not at the center) and of rooms like the Whispering Gallery, where a whisper at one point can be heard clearly at another point far away. Why does that happen? We are going to explain it here.

The ellipse as a locus

If you need to remember what "locus" means, play with this simpler case, the circle (one center, one distance):

Practice★★☆

Find the equation of the locus of 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…

Find the equation of the locus of 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 to obtain the standard equation of an ellipse.

Solve

Elements and standard equation

Elements of the ellipse (center at the origin, foci on the $X$-axis)

Semi-major axis $a$ and semi-minor axis $b$, with $a>b$. The major axis has length $2a$ and the minor axis $2b$.
Foci $F_1(-c,0)$ and $F_2(c,0)$, where $c^2=a^2-b^2$. The focal distance is $2c$.
Vertices: $(\pm a,0)$ on the major axis and $(0,\pm b)$ on the minor axis.
Standard equation: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.
Latus rectum: chord perpendicular to the major axis that passes through a focus; its length is $LR=\dfrac{2b^2}{a}$.

If the foci are on the $Y$-axis, the roles are swapped: the larger denominator goes under $y^2$ and the equation is $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$.

Why is $c^2=a^2-b^2$? At the vertex $(0,b)$, by symmetry, the distance to each focus is half of $2a$, that is, $a$. So the right triangle with legs $b$ and $c$ has hypotenuse $a$, and by Pythagoras $a^2=b^2+c^2$.

Practice★☆☆

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

Practice★☆☆

Write the standard equation of an ellipse centred 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…

Write the standard equation of an ellipse centred 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

Practice★☆☆

Find the equation of an ellipse centred at the origin, whose foci lie on the y-axis, knowing that the focal distance is $8$ units and the eccentricity is $0.8$.

Solve

Eccentricity: how flattened the ellipse is

Orbits and Kepler's laws
Johannes Kepler discovered that the planets move in ellipses with the Sun at one focus (first law). Also, the segment from the Sun to the planet sweeps out equal areas in equal times (second law): the planet moves faster when it is near the Sun. The minimum distance to the Sun is the perihelion, $a-c$, and the maximum is the aphelion, $a+c$. Mercury has the most eccentric orbit of the classical planets ($e\approx 0.21$); Earth's is almost circular ($e\approx 0.017$).

Eccentricity of an ellipse with semi-axes $a=10$ and $b=6$
Calculate the half focal distance: $c=\sqrt{a^2-b^2}=\sqrt{100-36}=\sqrt{64}=8$.
Apply the definition: $e=\dfrac{c}{a}=\dfrac{8}{10}$.
Write $e$ as a decimal and interpret the shape of the ellipse.
$e=0.8$: it is close to $1$, so the ellipse is quite elongated.

Orbit with known perihelion and aphelion
A satellite has a perihelion of $4$ units and an aphelion of $10$ (measured from the focus). Since $a-c=4$ and $a+c=10$, adding gives $2a=14$, so $a=7$.
Subtracting, $2c=6$ and $c=3$. Then $e=\dfrac{3}{7}\approx 0.43$ and $b=\sqrt{49-9}=\sqrt{40}\approx 6.32$.

What is the eccentricity of a circle?
Think about what happens to the foci when $a=b$.
Show answer
It is $e=0$: the two foci coincide at the center ($c=0$), so $e=c/a=0$.

Practice★☆☆Not started
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

Practice★☆☆Not started
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

Practice★★☆Not started
The orbit of Mercury around the Sun is an ellipse with the Sun at one of its foci. The minimum distance from Mercury to the Sun (perihelion) is $46$ million kilometers,…
The orbit of Mercury around the Sun is an ellipse with the Sun at one of its foci. The minimum distance from Mercury to the Sun (perihelion) is $46$ million kilometers, and the maximum distance (aphelion) is $70$ million kilometers. Calculate the eccentricity of Mercury's orbit and the length of its semi-minor axis in million kilometers.

Solve

Translated ellipses and the general equation

If the center is not at the origin, we do the same as with the circle: we replace $x$ with $x-h$ and $y$ with $y-k$. With center $C(h,k)$ and a horizontal major axis:
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$
If the major axis is vertical, $a^2$ goes under $(y-k)^2$. The vertices and foci shift just like the center: with a horizontal major axis, the foci are $(h\pm c,\,k)$.

Expanding the standard equation gives the general equation $Ax^2+Cy^2+Dx+Ey+F=0$, with $A$ and $C$ having the same sign and being different. To get back to the standard form we complete the square.

Complete the square in an orderly way
Group the terms in $x$ and in $y$ separately, factor out the coefficient of $x^2$ and of $y^2$, and when you complete the square add to the other side what you added, multiplied by that factor. At the end divide by the constant term so that it becomes $1$.

Converting $4x^2+9y^2-16x+18y-11=0$ to standard form
Group and factor: $4(x^2-4x)+9(y^2+2y)=11$.
Complete the square: $4\left[(x-2)^2-4\right]+9\left[(y+1)^2-1\right]=11$. So, $4(x-2)^2+9(y+1)^2=11+16+9=36$.
Divide by $36$ and give the center and the semi-axes.
$\dfrac{(x-2)^2}{9}+\dfrac{(y+1)^2}{4}=1$. Center $(2,-1)$, $a=3$, $b=2$, $c=\sqrt5$, foci $(2\pm\sqrt5,\,-1)$.

What is the center of $\dfrac{(x+5)^2}{9}+\dfrac{(y-3)^2}{4}=1$?
Regina says it is $(5,-3)$, copying the numbers from the parentheses.
Show answer
She has the signs wrong: the form is $(x-h)^2$, so $x+5=x-(-5)$ gives $h=-5$, and $k=3$. The center is $(-5,3)$.

Practice★☆☆Not started
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 lengths 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 lengths of its semi-axes.
Solve

Practice★☆☆Not started
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 canonical form.
Solve

Practice★★☆Not started
Given the general equation $9x^2 + 4y^2 - 36x + 24y + 36 = 0$, prove that it corresponds to an ellipse. To do so, express it in canonical 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 so, express it in canonical form and determine the coordinates of its center, its vertices, and its foci.
Solve

Points and lines relative to the ellipse

For a point $P(x_0,y_0)$ and the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, evaluate $\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}$:

if it equals $1$, the point is on the ellipse;

if it is less than $1$, it is inside (interior);

if it is greater than $1$, it is outside (exterior).

For a line, substitute its equation into that of the ellipse. You get a quadratic equation and its discriminant $\Delta$ decides: $\Delta>0$ the line is a secant (two points), $\Delta=0$ it is tangent (one point) and $\Delta<0$ it is exterior (no points).

Position of $x+y=3$ relative to $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
Clear the denominators: $4x^2+9y^2=36$. From the line, $y=3-x$.
Substitute: $4x^2+9(9-6x+x^2)=36$, that is $13x^2-54x+45=0$.
$\Delta=54^2-4\cdot13\cdot45=2916-2340=576>0$. The line is a secant. The solutions are $x=\dfrac{54\pm24}{26}$, that is $x=3$ and $x=\dfrac{15}{13}$; the points are $(3,0)$ and $\left(\dfrac{15}{13},\dfrac{24}{13}\right)$.

Is $(3,2)$ inside $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$?
Evaluate without drawing.
Show answer
$\dfrac{9}{25}+\dfrac{4}{9}=0.36+0.44\approx0.80<1$: yes, it is interior.

Practice★☆☆Not started
Check whether the points $P(2, -3)$, $Q(5, 0)$ and $R(4, 2)$ lie inside, outside, or on the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Check whether the points $P(2, -3)$, $Q(5, 0)$ and $R(4, 2)$ lie inside, outside, or on the ellipse with equation $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Solve

Practice★★☆Not started
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

Tangent lines

A line is tangent when it touches the ellipse at only one point. There are two ways.

With the discriminant. Write the line with a parameter (slope $m$ or intercept $k$), substitute it into the ellipse, and set $\Delta=0$.

With the point-of-tangency formula. If $(x_0,y_0)$ is on $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the tangent at that point is
$$\frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1.$$
For tangents from an exterior point, use the line $y-y_1=m(x-x_1)$ and set $\Delta=0$: you get two values of $m$.

Tangents to $x^2+4y^2=20$ of the form $x+2y+k=0$
Solve for $x=-2y-k$ and substitute: $(2y+k)^2+4y^2=20$, that is $8y^2+4ky+k^2-20=0$.
Set $\Delta=0$: $16k^2-32(k^2-20)=0$, so $-16k^2+640=0$ and $k^2=40$.
Write the two tangent lines.
$k=\pm2\sqrt{10}$: the lines are $x+2y+2\sqrt{10}=0$ and $x+2y-2\sqrt{10}=0$.

A tangent through an exterior point gives two lines
When you set $\Delta=0$ you get a quadratic equation in $m$. Two slopes come out (two tangents). If you give only one, the other is missing; and if $\Delta<0$, the point is not exterior.

How many tangents to an ellipse pass through an interior point?
Think about what happens if the line already cuts the ellipse at two points.
Show answer
None: any line through an interior point cuts the ellipse at two points, so it is never tangent.

Practice★★★Not started
Calculate the values of the parameter $k$ such 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 $k$ such that the line $x + 2y + k = 0$ is tangent to the ellipse with equation $x^2 + 4y^2 = 20$.
Solve

Practice★★★Not started
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

Practice★★★Not started
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

Problems and geometric loci

Many problems are solved in two steps: identify $a$, $b$ and $c$, and write the equation so you can substitute data.

An elliptical arch
A tunnel has a semi-elliptical arch $12\text{ m}$ wide and $4\text{ m}$ high. What is its height at $3\text{ m}$ from the center?
With the center at the origin, $a=6$ and $b=4$: $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$.
For $x=3$: $\dfrac{y^2}{16}=1-\dfrac{9}{36}=\dfrac34$, so $y^2=12$ and $y=2\sqrt3\approx3.46\text{ m}$.

The whispering gallery. An elliptical room works because any sound leaves one focus, bounces off the wall, and arrives at the other focus: all the paths have the same length, $2a$, and so they arrive together and reinforce each other.

Other characterizations. The ellipse is also the locus of points whose distance to a focus is $e<1$ times their distance to a fixed line (the directrix). And if you "squash" a circle $x^2+y^2=r^2$ by multiplying the ordinates by a factor $k$ with $0
Locus with a constant ratio
The points $P(x,y)$ whose distance to $F(2,0)$ is half their distance to the line $x=8$:
$\sqrt{(x-2)^2+y^2}=\dfrac12|x-8|$. Square both sides: $4[(x-2)^2+y^2]=(x-8)^2$.
Expand: $4x^2-16x+16+4y^2=x^2-16x+64$, so $3x^2+4y^2=48$, that is $\dfrac{x^2}{16}+\dfrac{y^2}{12}=1$. It is an ellipse with $a=4$, $b=2\sqrt3$, $c=2$ and $e=\tfrac12$.

Internally tangent circles. Two circles are internally tangent when the small one (center $C$, radius $r$) touches the large one (center $O$, radius $R$) at exactly one point. Then the distance between centers is the difference of the radii: $d(O,C)=R-r$. If in addition the small one passes through a fixed point $A$, its radius is $r=d(C,A)$. Substituting, $d(O,C)+d(C,A)=R$: a constant sum of distances to two fixed points, that is, an ellipse with foci $O$ and $A$.

Centers of circles internally tangent to $x^2+y^2=64$ that pass through $A(4,0)$
The large one has center $O(0,0)$ and $R=8$. Let $C(x,y)$ be the center of the small one: $d(O,C)=8-r$ and $r=d(C,A)$.
So $d(C,O)+d(C,A)=8$. It is an ellipse with foci $O(0,0)$ and $A(4,0)$, with $2a=8$, that is $a=4$.
The focal distance is $2c=4$, so $c=2$ and $b^2=16-4=12$. The center of the ellipse is the midpoint of the foci, $(2,0)$. The equation is $\dfrac{(x-2)^2}{16}+\dfrac{y^2}{12}=1$.

What conic do you get if the ratio of distances (focus / directrix) is $\tfrac12$?
Recall the range of the eccentricity of each conic.
Show answer
An ellipse: the ratio is the eccentricity, and $0

Practice★★☆Not started
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 the height of the arch at a distance of $4\text{ m}$ from the center of the river.

Solve

Practice★★☆Not started
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 should each person stand to take advantage of this acoustic phenomenon?
Solve

Practice★★☆Not started
Determine the locus of 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 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 section and calculate its center and the lengths of its semi-axes.
Solve

Practice★★☆Not started
In a clockwork mechanism, a gear has an elliptical shape. It is known that its boundary equation is given by $x^2 + 4y^2 = 16$. A small pin is to be placed at a point on…
In a clockwork mechanism, a gear has an elliptical shape. It is known that its boundary equation is given by $x^2 + 4y^2 = 16$. A small pin is to be placed 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

Practice★★★Not started
Consider the circle with equation $x^2 + y^2 = 100$. With each point $A(x, y)$ on the circle, we associate a point $P(x, y')$ that has the same x-coordinate, but its…
Consider the circle with equation $x^2 + y^2 = 100$. With each point $A(x, y)$ on the circle, we associate 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

Practice★★★Not started
A rectangular window is to be designed, 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, 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 required to have a total area of $24\text{ m}^2$, what should its exact dimensions be?
Solve

Practice★★★Not started
A main square is shaped like an ellipse. A straight path is to be paved connecting the leftmost focus, $F_1$, with a point $P$ on the ellipse located in the second…
A main square is shaped like an ellipse. A straight path is to be paved connecting the leftmost focus, $F_1$, with a point $P$ on the ellipse located in the second quadrant. It is known that the major axis of the square measures $100\text{ m}$ and the minor axis measures $60\text{ m}$. If the angle formed by the segment $F_1P$ with the positive x-axis is $60^\circ$, calculate the exact length of the paved path.

Solve

Practice★★★Not started
An ellipse is centred at the origin and its axes of symmetry coincide with the Cartesian coordinate axes. Given that the ellipse passes through the points…
An ellipse is centred at the origin and its axes of symmetry coincide with the Cartesian coordinate axes. Given 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

Practice★★★Not started
Find the locus of the centers of the circles that are internally tangent to the circle with equation $x^2 + y^2 = 100$ and pass through the fixed point $A(6, 0)$. Prove…
Find the locus of the centers of the circles that are internally tangent to the circle with equation $x^2 + y^2 = 100$ and pass through the fixed point $A(6, 0)$. Prove that this locus is an ellipse, find its canonical equation and determine the coordinates of its foci.
Solve

Back to the beginning: in the whispering gallery, someone whispering at one focus is heard at the other, because all the bounces travel the same total distance $2a$ and the waves arrive at the same time.

Let's practice! 8 medium exercises in this section
★★☆ The orbit of Mercury around the Sun is an ellipse with the Sun at one of its foci. The…Not started ★★☆ A stone bridge over a river has the shape of a semi-elliptical arch. The width of the…Not started ★★☆ Given the general equation $9x^2 + 4y^2 - 36x + 24y + 36 = 0$, prove that it corresponds…Not started

Summary

Summary

Ellipse: $d(P,F_1)+d(P,F_2)=2a$. Standard equation with center at the origin: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$; the larger denominator indicates the major axis).

$c^2=a^2-b^2$; foci $(\pm c,0)$; latus rectum $\dfrac{2b^2}{a}$; eccentricity $e=\dfrac{c}{a}$, with $0

With center $(h,k)$: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$. The general equation is converted to this form by completing the square.

Point: evaluate the left side ($=1$ on the ellipse, $<1$ inside, $>1$ outside). Line: substitute and look at the discriminant ($>0$ secant, $=0$ tangent, $<0$ exterior).

Circles internally tangent to another circle and passing through a fixed point: their centers satisfy $d(O,C)+d(C,A)=R$, an ellipse.

In orbits the astronomical body is at a focus: perihelion $a-c$, aphelion $a+c$.

Let's practice! 8 hard exercises in this section
★★★ Find the equations of the tangent lines to the ellipse with equation…Not started ★★★ Determine the equations of the tangent lines to the ellipse $x^2 + 2y^2 = 18$ that are…Not started ★★★ Calculate the values of the parameter $k$ such that the line $x + 2y + k = 0$ is tangent…Not started ★★★ Consider the circle with equation $x^2 + y^2 = 100$. With each point $A(x, y)$ on the…Not started

See all the exercises in this section
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