﻿ Ellipse calculator
 Ellipse calculator ▲ Polar form when the left focus point is at the origin:
Ellipse equation
 x2 + y2 = 1 2 2
 x2 + y2 =
Semi-major axis (a)
Semi-minor axis (b)
Area (A)
Accentricity (e)
Foci distance (c) ±
Flattening factor (g)
vertices ±
Ellipse center
Circumference (P)
 Input limit:
 NOTE: The perimeter of the ellipse is calculated by using accurate solution of infinite series to the accuracy selected. The increase of accuracy or the ratio a/b causes the calculator to use more terms to reach the selected accuracy.
The general ellipse equation with center not at (0,0) is:
 x2 + y2 + x + y + = 0
 (x − )2 + ( y − )2 = 1 2 2
 Ellipse ▲ 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. If   b > a   then the ellipse is a vertical ellipse and the foci are on the  y  axis.
 In this case the equation of the ellipse is: 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 slop 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:
 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  From ellipse definition   d1 + d2 = 2a From the drawing   d1 = d2 2d1 = 2a And we get the relation   d1 = a   and: The points  A1  and  A2  in this case  (±a , 0)  are the vertices of the major axis.
 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 - distances from foci ▲
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 rearanging terms is: 3x2 + 4y2 − 36x = 0