Polar form: 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 location

Ellipse center

Perimeter (P)

Input limit:

Notes:

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.

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 a horizontal ellipse is:

(center at x = 0 y = 0)

Equation of a vertical ellipse is:

(Center at x = 0 y = 0)

Notice that the values of a and b has swapped, this was done in order to keep
the value of the vertex as a (see drawing). Keep in mind that in order to calculate the slope of the
tangent line do not take the swapped values.

The eccentricity e of an ellipse is:

(a > b)

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:

Notes:(1) When a = b then f = 0 it means that the ellipse is a circle.

(2) In vertical ellipse the values of a and b should be swapped.

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

( b > a ) vertices are at:

(0 , a) and (0 , −a )

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.

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

Note: In this equations do not change between a and b even if b > a

If the center of the ellipse is moved by x = h and y = k then if a point on the ellipse is given the corresponding x or y coordinate is calculated by the equations:

Ellipse equation

Given

Equivalent point on ellipse

x_{1}

y_{1}

Notice: these equations are good for horizontal and vertical ellipses.

Ellipse

Center

(h , k)

(h , k)

Vertices

(h − a , k) (h + a , k)

(h , k − a) (h , k + a)

Foci

(h − c , k) (h + c , k)

(h , k − c) (h , k + c)

The slope of the line tangent to the ellipse at point (x1 , y1) is:

The equation of the tangent line at point (x1 , y1) on the ellipse is:

Find the equation of the ellipse that has eccentricity of 0.75, and the foci is along the x axis and second time if the foci is along the 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):

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:

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:

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 this 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 15 − distance from a point on ellipse to foci

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.

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: