Ellipse with center at
(x
_{1}
, y
_{1}
)
calculator
▲
x
^{2}
+
y
^{2}
+
x
+
y
+
= 0
(x
−
)
^{2}
+
( y
−
)
^{2}
= 1
2
2
Polar form when the left focus point is at the origin:
Semi-major axis (a)
Semi-minor axis (b)
Area (A)
Eccentricity (e)
Foci length (c)
Flattening factor (f)
Foci location
vertices ±
Ellipse center
Perimeter (P)
Input limit:
Notes:
Ellipse summary
Ellipse equation
Ellipse tangent line
Vertices and eccentricity
Foci and eccentricity
Ellipse center at (x,y)
Ellipse equation
Ellipse equation forms
Area of ellipse
Ellipse tangent lines 1
Ellipse tangent lines 2
Minor and major axes
Ellipse and tangent points
Distance from foci
Ellipse summary
▲
An ellipse is the locus of all points that the sum of whose distances from two fixed points is constant,
d
_{1}
+ d
_{2}
= constant = 2a
the two fixed points are called the foci (or in single focus).
distance between both foci is: 2c
a and b − major and minor radius
Equation of an ellipse:
(center at x = 0 y = 0)
The
eccentricity
e of an ellipse:
Where (c = half distance between foci) c < a 0 < e < 1
If
e = 0
then the ellipse is a circle.
If
0 < e < 1
then it is an ellipse.
If
e = 1
then the ellipse is a parabola.
If
e > 1
then the ellipse is a hyperbola.
The
flattening
f of an ellipse is the amount of the compression of a circle along a diameter to
form an ellipse and its value is:
Note that when a = b then f = 0 it means that the ellipse is a circle.
The
vertices
of an ellipse are the intersection points of the major axis and the ellipse. The line segment joining the vertices is the major axis, and its midpoint is the center of the ellipse.
For a horizontal ellipse
( a > b ) vertices are at:
(a , 0) and (−a , 0)
For a vertical ellipse
( a < b ) vertices are at:
(0 , b) and (0 , −b )
From ellipse definition d + d = const
And from x direction 2c + 2(a − c) = const
2d = 2a
And we get the relation d = a and:
The points A
_{1}
and A
_{2}
in this case (±a , 0) are the vertices of the major axis.
Example:
Given the ellipse with equation 81x
^{2}
+ 4y
^{2}
= 324
find ellipse parameters.
Solution
: Divide by 324. to obtain
Since a < b ellipse is vertical with foci at the y axis and a = 9 and b = 2.
Eccentricity:
Flattening:
Focus c:
c = a * e = 8.775
Area:
A = π2∙9 = 56.55
The slope of the tangent line to the ellipse at point (x
_{1}
, y
_{1}
) is:
The tangent line equation at a point (x
_{1}
, y
_{1}
) on the ellipse
or
From ellipse equation:
The area of an ellipse:
(see integral)
The integral value is:
If the center of the ellipse is moved by x = h and y = k then the equations of the ellipse become:
Ellipse center is at (h, k):
If a point y
_{1}
is given then:
If a point x
_{1}
is given then:
Converting ellipse presentation formats:
(See detail calculation)
①
Ax
^{2}
+ By
^{2}
+ Cx + Dy + E = 0
②
① → ②
Define:
② → ①
A = b
^{2}
B = a
^{2}
C = − 2hb
^{2}
D = − 2ka
^{2}
E = a
^{2}
k
^{2}
+ b
^{2}
h
^{2}
− a
^{2}
b
^{2}
Any point from the center to the circumference of the ellipse can be expressed by the angle θ in the
range (0 − 2π) as:
x = a
cos
θ y = b
sin
θ
If we substitute the values x = r
cos
θ and y = r
sin
θ in the equation of the ellipse we can get the
distance of a point from the center of the ellipse r(θ) as:
If the origin is at the left focus then the ellipse equation is:
The perimeter (P) of an ellipse is found by integration:
The only solution is by series:
Where e is the eccentricity of the ellipse
Another solution is by using the series:
Where
Ramanujan approximation for the circumference:
Where
Less accurate approximation
Verify the equation of an ellipse
▲
From the definition of the ellipse, we know that d
_{1}
+ d
_{2}
= 2a
Where a is equal to the x axis value or half the major axis.
From the drawing d
_{1}
and d
_{2}
are equal to:
Simplify the equation by transferring one radical to the right and squaring both sides:
After rearranging terms, we obtain:
We square again both sides to find:
After arranging terms, we get:
Dividing by a
^{2}
− c
^{2}
we find:
Since a > c we can introduce a new quantity:
And the equation of an ellipse is revealed:
If the foci are placed on the y axis then we can find the equation of the ellipse the same way: d
_{1}
+ d
_{2}
= 2a
Where a is equal to the y axis value or half the vertical axis.
From the drawing d
_{1}
and d
_{2}
are equal to:
After arranging terms and squaring we get:
After rearranging terms, we find:
Set new quantity (see above):
Dividing by b
^{2}
we get the final form:
Example - Tangent line to an ellipse
▲
Find the equation of the line tangent to the ellipse
4x
^{2}
+ 12y
^{2}
= 1
at the point P(0.25 , 0.25).
By implicit differentiation we will find the value of dy/dx that is the slope at any x and y point.
Implicit differentiation dy/dx is:
The value of dy/dx is:
At the given point the slope is:
Equation of the tangent
line that passes through the point P and has slope m
is: y = mx + ( y
_{p}
− mx
_{p}
)
Substitute the point P(0.25 , 0.25) we get:
y = mx + (0.25 - m * 0.25)
And the tangent line equation is:
x + 3y − 1 = 0
Example - vertices and eccentricity
▲
Find the equation of the ellipse that has vertices at (0 , ± 10) and has eccentricity of 0.8.
Notice that the vertices are on the y axis so the ellipse is a vertical ellipse and we have to use the vertical ellipse equation.
The equation of the eccentricity is:
After multiplying by a we get:
e
^{2}
a
^{2}
= a
^{2}
− b
^{2}
The value of b
^{2}
is:
b
^{2}
= a
^{2}
(1 − e
^{2}
)
The equation of a vertical ellipse is:
And the final equation of the ellipse is:
Example - foci and eccentricity
▲
Find the equation of the ellipse that has eccentricity of 0.75, and the foci along 1. x axis 2. y axis, ellipse center is at the origin, and passing through the point (6 , 4).
The point (6 , 4) is on the ellipse therefore fulfills the ellipse equation.
1.
Substitute the point (x
_{1}
, y
_{1}
) into the ellipse
equation (foci at x axis):
From the previous example:
b
^{2}
= a
^{2}
(1 − e
^{2}
)
Substitute b
^{2}
into ellipse equation:
The value of a
^{2}
is:
The value of b
^{2}
is:
And the equation of the ellipse is:
7x
^{2}
+ 16y
^{2}
= 508
2.
Vertical ellipse equation is (foci at y axis):
Substitute b
^{2}
into ellipse equation:
The value of a
^{2}
is:
The value of b
^{2}
is:
And the equation of the ellipse is:
16x
^{2}
+ 7y
^{2}
= 688
Example - Translated center of ellipse
▲
Find the vertices and the foci coordinate of the ellipse given by
3x
^{2}
+ 4y
^{2}
- 12x + 8y + 4 = 0
.
Find the square in x and y:
3(x
^{2}
- 4x ) + 4(y
^{2}
+ 2y ) = − 4
Add and subtract 4 to the left parentheses and 1 to the right parentheses to obtain:
3(x − 2)
^{2}
− 12 + 4(y + 1)
^{2}
− 4 = − 4
3(x − 2)
^{2}
+ 4(y + 1)
^{2}
= 12
After dividing by 12 we get:
The center of this ellipse is at (2 , − 1) h = 2 and k = − 1.
Translate the ellipse axes so that the center will be at (0 , 0) by defining:
x' = x − 2
y' = y + 1
now the ellipse equation in the x'y' system is:
Which we recognize as an ellipse with vertices a = ± 2
and the foci is
In the xy system we have the vertices at (2 ± 2 , − 1) and the foci at (2 ± 1 , − 1).
The sketch of the ellipse is:
Example - Ellipse center
▲
Find the equation of the locus of all points the sum of whose distances from (3, 0) and (9, 0) is 12.
It can be seen that the foci are lying on the line y = 0 so the ellipse is horizontal.
The focus is equal to:
From the definition of the ellipse, we know that:
d
_{1}
+ d
_{2}
= 2a
and the value of a is 12 = 2a
a = 6
and the value of minor vertex b is:
The ellipse in the x'y' system is:
The ellipse in the xy system is:
The ellipse after rearranging terms is:
3x
^{2}
+ 4y
^{2}
− 36x = 0
Converting ellipse presentation formats
▲
Find the equation of the translation between the two forms of ellipse presentation.
①
Ax
^{2}
+ By
^{2}
+ Cx + Dy + E = 0
②
From equation ② we have:
b
^{2}
( x − h )
^{2}
+ a
^{2}
( y − k )
^{2}
= a
^{2}
b
^{2}
b
^{2}
( x
^{2}
- 2hx +h
^{2}
) + a
^{2}
( y
^{2}
− 2ky + k
^{2}
) = a
^{2}
b
^{2}
And finally:
b
^{2}
x
^{2}
- 2b
^{2}
hx + b
^{2}
h
^{2}
+ a
^{2}
y
^{2}
− 2a
^{2}
ky + a
^{2}
k
^{2}
− a
^{2}
b
^{2}
= 0
After rearranging by powers:
b
^{2}
x
^{2}
+ a
^{2}
y
^{2}
− 2b
^{2}
h
x
− 2a
^{2}
k
y
+ b
^{2}
h
^{2}
+ a
^{2}
k
^{2}
− a
^{2}
b
^{2}
= 0
Now we can find the values of the coefficients of the ellipse equation ① A, B, C, D and E.
A = b
^{2}
B = a
^{2}
C = − 2b
^{2}
h
D = − 2a
^{2}
k
E = b
^{2}
h
^{2}
+ a
^{2}
k
^{2}
− a
^{2}
b
^{2}
The transformation from equation ② to equation ① includes more steps to solve:
From equation ① we have:
(A
x
^{2}
+ C
x
) + (B
y
^{2}
+ D
y
) = −E
Take A and B out of both parenthesis:
Now we use the square formula of the form x
^{2}
+ 2Rx + R
^{2}
= (x + R)
^{2}
to get the square equations:
We have to add the following values to the right side of the equation:
(3)
In order to simplify the equation, we set:
We get:
A
(
x −h
)
^{2}
+
B
(
y −k
)
^{2}
= − E + A h
^{2}
+ B k
^{2}
Simplify again by setting the value: φ = − E + A h
^{2}
+ B k
^{2}
to get the equation:
A
(
x −h
)
^{2}
+
B
(
y −k
)
^{2}
= φ
(4)
Divide equation
(4)
by AB:
(5)
Divide both sides of equation
(5)
by the value of the right side of
(5)
φ/AB the result is:
We got the equation of the ellipse where h and k are the center of the ellipse and the denominators are the square values of the semi major and minor length a
^{2}
and b
^{2}
the complete transformation are:
Example - area of an ellipse
▲
Find the area of an ellipse if the length of major axes is 7 and the length of minor axes is 4
Example - tangent lines to ellipse 1
▲
Find the slope and the tangent line equation at a point where x
_{1}
= 2 on the ellipse
The general equation of an ellipse with center at (0 , 0) is:
Implicit differentiation of the ellipse equation relative to x:
Eliminating
= m (slope) from the derivation yields:
Line equation at point (x
_{1}
, y
_{1}
)
is:
y − y
_{1}
= m
(
x − x
_{1}
)
(1)
Substitute the value of m (slope of the line dy/dx) into equation
(1)
to get the equation of the line tangent to the ellipse at point (x
_{1}
, y
_{1}
):
Multiplying all terms by a
^{2}
y
_{1}
a
^{2}
y y
_{1}
− a
^{2}
y
_{1}
^{2}
=
−b
^{2}
x x
_{1}
+ b
^{2}
x
_{1}
^{2}
(2)
Point (x
_{1}
, y
_{1}
) is on the ellipse therefore it satisfies the ellipse equation:
(3)
Substitute eq (3) into eq (2) we get the general form of a tangent line to an ellipse at point
(
x
_{1}
, y
_{1}
)
b
^{2}
x
_{1}
x
+
a
^{2}
y
_{1}
y − a
^{2}
b
^{2}
= 0
(4a)
Or after dividing by a
^{2}
y
_{1}
we get:
(4b)
Now we should find the tangent points where x
_{1}
= 2
therefore substitute the point x
_{1}
in the equation of the ellipse and eliminate y we should get two values for y corresponding to the upper and lower sides of the ellipse (see sketch below).
The slopes of the tangent lines are:
Notice that when point y
_{2}
is negative then the slope m is positive and vies versa.
from eq
(
4b
)
the equations of the tangent lines are:
y
=
− 0.43 x
+
3.46
y
=
0.43 x − 3.46
Example - tangent lines to ellipse 2
▲
Find the equation of the line tangent to the ellipse 4x
^{2}
+
12y
^{2}
= 1
at the point P
(0.25 , 0.25)
.
The slope of the tangent line
can be found by implicit derivation of the ellipse equation:
Eliminate dy/dx to obtain:
The slope at point x = 0.25 is:
(slope of the tangent line)
The tangent line equation at the given point is:
y − y
_{1}
=
m
(
x − x
_{1}
)
Substitute given values for x
_{1}
and y
_{1}
:
And finally tangent line equation is:
x + 3y − 1 = 0
Example - foci and minor and major axes
▲
Given the ellipse 4x
^{2}
+
9y
^{2}
− 16x
+
108y
+
304
= 0
find the lengths of the minor and major axes, the coordinates of the foci and eccentricity.
Completing the square for both x and y we have
(
4x
^{2}
− 16x
) + (
9y
^{2}
+
108y
) +
304
= 0
(
2x − 4
)
^{2}
− 16
+ (
3y
+
18
)
^{2}
− 324
+ 304 = 0
4
(
x − 2
)
^{2}
+
9
(y +
6
)
^{2}
= 36
Divide by 36:
which tells us that a = 3, b = 2
Since a
^{2}
= b
^{2}
+ c
^{2}
we find that
and the focus coordinates on the x axis are:
The eccentricity (only the positive value) is:
Notice that a, b, h and k can be found by using the equations that had been derived earlier:
Substituting all values to the equation of the ellipse we get:
or
Example - distance of a line from ellipse
▲
Given the ellipse 16x
^{2}
+
25y
^{2}
=
400 and the line y
=
−x
+
8 find the minimum and maximum distance from the line to the ellipse and the equation of the tangents lines.
Divide the ellipse equation by 400 to get the general form of the ellipse, we can see that the major and minor lengths are a = 5 and b = 4:
The slope of the given line is m
=
−
1
this slope is also the slope of the tangent lines that can be written by the general equation y = −x
+ c
(c is a constant).
Because the tangent point is common to the line and ellipse we can substitute this line equation into the ellipse equation to get:.
Completing the square for both x and y we have
(
4x
^{2}
− 16x
) + (
9y
^{2}
+
108y
) +
304
= 0
41x
^{2}
− 50cx + 25c
^{2}
− 400 = 0
And the solution of the square equation is:
Notice that two different solutions for x will give us intersection of an ellipse and a line therefore we need only one solution for tangency condition that will happen when the expression under the root will be equal to 0.
65600 − 1600c
^{2}
= 0
and
The general tangent line equation is:
y
=
mx
+
c
substitute m = − 1 and c
_{1,2}
into this equation will give us the two tangent lines which are:
x
+
y
+
6.4
= 0
and
x
+
y − 6.4
= 0
Distances d and D (see drawing) are the distances between the tangency lines and the given line and can be found according to the equation for the
distance between two lines
:
where line equations are:
Ax + By + C = 0
and
Dx + Ey +F = 0
In our case A = B = C = 1 so the distance reduces to:
Example - distance from a point on ellipse to foci
▲
Find the points on the ellipse
whose distance from the right foci is 6.
Foci values are:
Foci coordinates are:
(− 4 , 0) and (4 , 0)
From ellipse definition we have:
d
_{1}
+ d
_{2}
= 2c + 2(a − c) = 2a
After substitute d
_{2}
= 5 we get d
_{1}
= 2a − 6 = 10 − 6 = 4
Points (x
_{1}
, y
_{1}
) and (x
_{2}
, y
_{2}
) can be found geometrically.
x
_{1}
=
c − d
_{2}
cos
θ
(1)
The value of
cos
θ can be calculated from
cos
law
:
d
_{1}
^{2}
=
d
_{2}
^{2}
+
4c
^{2}
− 2
cos
θ 2cd
_{2}
(2)
From eq.
(1)
x
_{1}
= 4 − 6 · 0.875 = −
1.25
y
_{1}
= d
_{2}
sin
θ = 6 • 0.484 = 2.9
The points on ellipse that are 6 units from the foci are:
(
−
1.25 , 2.9)
(
−
1.25 , −
2.9
)
The answer can be checked by calculating the distance between the calculated point and the foci.
Another way to solve the problem is to find the intersection points of a circle whose radius is d
_{2}
and with center at the right foci and the given ellipse.
(x − c)
^{2}
+ y
^{2}
= d
_{2}
^{2}
From ellipse equation we have:
Substitute y
^{2}
into the circle equation we get:
We get a
quadratic equation
for the x coordinate:
0.64x
^{2}
−
8
x −
11 = 0
The solutions of this equation are: −
1.25
and 13.75 the second solution is located outside the ellipse so the only solution is x
_{1}
= −
1.25
as expected.
The value of y coordinate can be calculated from the ellipse equation:
NOTES
(1)
The perimeter of the ellipse is calculated by using infinite series to the selected accuracy.
The increase of accuracy or the ratio a / b causes the calculator to use more terms to reach the selected accuracy. Present calculation used:
iterations.
(2)
Notice that pressing on the sign in the equation of the ellipse or entering a negative number changes the + / − sign and changes the input to positive value.