﻿ Ellipse calculator
 Ellipse with center at (x1 , y1) calculator ▲
 x2 + y2 + 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)
Accentricity (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 accentricity 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,
d1 + d2 = 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  A1  and  A2  in this case  (±a , 0)  are the vertices of the major axis.
Example: Given the ellipse with equation    81x2 + 4y2 = 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 (x1 , y1) is:
The tangent line equation at a point   (x1 , y1)   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 ellips become:
 Ellipse center is at (h, k):
 If a point   y1   is given then: If a point   x1   is given then:
Converting ellipse presentation formats: (See detail calculation)
 ① Ax2 + By2 + Cx + Dy + E = 0 ②
 ① → ② Define:
 ② → ① A = b2 B = a2 C = − 2hb2 D = − 2ka2 E = a2k2 + b2h2 − a2b2
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 cosθ               y = b sinθ
If we substitute the values   x = r cosθ   and   y = r sinθ   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 equstion 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     d1 + d2 = 2a
Where  a  is equal to the x axis value or half the major axis.
From the drawing d1 and d2 are equal to:
Simplify the equation by transferring one redical 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   a2 − c2   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:   d1 + d2 = 2a
Where  a  is equal to the y axis value or half the vertical axis.
From the drawing d1 and d2 are equal to:
 After arranging terms and squaring we get: After rearranging terms we find: Set new quantity (see above): Dividing by b2 we get the final form:
 Exampe - Tangent line to an ellipse ▲
Find the equation of the line tangent to the ellipse   4x2 + 12y2 = 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 + ( yp − mxp) 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 multipling by a we get: e2a2 = a2 − b2 The value of   b2   is: b2 = a2(1 − e2) 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 accentricity 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 (x1 , y1) into the ellipseequation (foci at x axis): From the previous example: b2 = a2(1 − e2) Substitute   b2   into ellipse equation: The value of   a2   is: The value of   b2   is: And the equation of the ellipse is: 7x2 + 16y2 = 508 2. Vertical ellipse equation is (foci at y axis): Substitute   b2   into ellipse equation: The value of   a2   is: The value of   b2   is: And the equation of the ellipse is: 16x2 + 7y2 = 688
 Example - Transelated center of ellipse ▲
Find the vertices and the foci coordinate of the ellipse given by     3x2 + 4y2 - 12x + 8y + 4 = 0.
 Find the square in x and y: 3(x2 - 4x       ) + 4(y2 + 2y       ) = − 4 Add and subtruct 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: d1 + d2 = 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: 3x2 + 4y2 − 36x = 0
 Converting ellipse presentation formats ▲
Find the equation of the translation between the two forms of ellipse presentation.
 ① Ax2 + By2 + Cx + Dy + E = 0 ②
 From equation ② we have: b2 ( x − h )2 + a2 ( y − k )2 = a2b2 b2 ( x2 - 2hx +h2) + a2( y2 − 2ky + k2) = a2b2 And finally: b2 x2 - 2b2hx + b2h2 + a2y2 − 2a2ky + a2k2 − a2b2 = 0 After rearranging by powers: b2 x2 + a2 y2 − 2b2h x − 2a2k y + b2h2 + a2k2 − a2b2 = 0
Now we can find the values of the coefficients of the ellipse equation   ①   A, B, C, D and E.
 A = b2 B = a2 C = − 2b2h D = − 2a2k E = b2h2 + a2k2 − a2b2
 The transformation from equation ② to equation ① includes more steps to solve: From equation ① we have: (A x2 + C x) + (B y2 + D y) = −E
 Take A and B out of both parenthesis:
Now we use the square formula of the form     x2 + 2Rx + R2 = (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 h2 + B k2
Simplify again by setting the value:           φ = − E + A h2 + B k2           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  a2  and  b2  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  x1 = 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  (x1 , y1)  is: y − y1 = m (x − x1)
(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  (x1 , y1):
Multiplying all terms by  a2 y1 a2 y y1 − a2 y12 = −b2 x x1 + b2 x12 (2)
Point (x1 , y1) 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 (x1 , y1)
b2 x1 x + a2 y1 y − a2b2 = 0 (4a)
Or after dividing by  a2 y1  we get: (4b)
Now we should find the tangent points where  x1 = 2  therfore substitute the point  x1  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  y2  is negative then the slope m  is positive and vise 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  4x2 + 12y2 = 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 − y1 = m (x − x1) Substitute given values for  x1  and  y1: And finally tangent line equation is: x + 3y − 1 = 0
 Example - foci and minor and major axes ▲
Given the ellipse   4x2 + 9y2 − 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 (4x2 − 16x) + ( 9y2 + 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   a2 = b2 + c2     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   16x2 + 25y2 = 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 elipse 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 ia 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 (4x2 − 16x) + ( 9y2 + 108y) + 304 = 0 41x2 − 50cx + 25c2 − 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 therfore we need only one solution for tangency condition that will happen when the expression under the root will be equal to 0.
 65600 − 1600c2 = 0 and
The general tangent line equation is: y = mx + c
substitute  m = − 1  and  c1,2  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: d1 + d2 = 2c + 2(a − c) = 2a After substitute  d2 = 5  we get  d1 = 2a − 6 = 10 − 6 = 4 Points  (x1 , y1)  and  (x2 , y2)  can be found geometrically.
 x1 = c − d2 cosθ (1)
The value of  cosθ  can be calculated from  cos  law:
 d12 = d22 + 4c2 − 2 cosθ 2cd2 (2)
 From eq. (1) x1 = 4 − 6 · 0.875 = −1.25 y1 = d2 sinθ = 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 d2 and with center at the right foci and the given ellipse.
 (x − c)2 + y2 = d22 From ellipse equation we have: Substitute   y2   into the circle equation we get: We get a quadratic equation for the x coordinate: 0.64x2 −8x −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   x1 = −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.