﻿ AmBrSoft.net Ellipse calculator
 Ellipse with center at (h , k) calculator ▲
 x2 + y2 + x + y + = 0
 (x − )2 + ( y − )2 = 1 2 2
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:
 Line tangent to ellipse
 A point on the ellipse x1   y1
 Tangent line slope   m = tanθ: Line angle θ :
 Tangency point Tangent line equation Line angle Point 1 Point 2
 Summary center (0,0)  (h,k) 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 Tangent line
 Ellipse summary − center at  (0 , 0) ▲
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 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  A1  and  A2  in this case  (±a , 0)  are the vertices of the major axis.
 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
Note: In this equations do not change between a and b even if  b > a
The area of an ellipse is exactly:
A = πab
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
 Ellipse summary − center at  (h , k) ▲
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
x1
y1
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:

 Or:
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
Polar coordinate of ellipse:
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 equation is:
 1 − 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 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   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:
 Example 2 − Tangent line to an ellipse with center at (h , k) ▲
Find the equations of the tangent lines at the points of the intersection of line  x = 5.5  and the ellipse
 First we will solve the general case of the tangent line to the ellipse.
 The slope of the tangent line by explicit derivation: And the slope value at a point on the ellipse is: The tangent line equation at point (x1 , y1) is: mx − y + (y1 − mx1) = 0
 After rearranging terms: b2 (x1 + h) x + a2 (y1 + k) y − a2 (y1 + k) y1 − b2 (x1 + h) x1 = 0
Notice that if the ellipse is located in the origin than  h = k = 0  and the line equation turns to be
b2x1x + a2y1y − a2y12 − b2x12 = 0

Now we will find the two points  y1,2  of the intersection of the given line  x = 5.5  with the ellipse.

 9 (5.5 − 4)2 + 3(y + 5)2 = 27 After solving for  y  we get The intersection points are: (5.5 , − 3.5)         (5.5 , − 6.5)

Now we have to find the slope of the tangent line to the ellipse by implicit derivation.

 After derivation we get: And the slope m is:

The slopes of the tangent lines at the two tangent points are:

Now we can find the equation of the tangents lines according to slope and a point.

 First line: mx - y + y1 − mx1 = 0 → -3x - y − 3.5 + 3 * 5.5 = 0 → 3x + y − 13 = 0 Second line: mx - y + y2 − mx2 = 0 → 3x - y − 6.5 + 3 * 5.5 = 0 → 3x − y − 23 = 0
 Example 3 − 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 4 − vertices and eccentricity ▲
Given an ellipse by the equation  81x2 + 4y2 = 324  find the eccentricity, flattening, area and the focus distance c of the ellipse.
 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
 Example 5 − 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: 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 6 − foci and eccentricity ▲
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 (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 7 − Translated center of ellipse ▲
Find the vertices and the foci coordinate of the ellipse given by     3x2 + 4y212x + 8y + 4 = 0.
 Find the square in x and y: 3(x2 - 4x       ) + 4(y2 + 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 8 − 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
 9 − 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 10 − 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 11 − tangent lines to ellipse ▲
 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  therefore 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 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 12 − tangent lines to ellipse ▲
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 13 − foci and minor and major axes ▲
Given the ellipse   4x2 + 9y216x + 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

From the ellipse equation we see that the center of the ellipse is at:     (2 ,6)

 Since   a2 = b2 + c2     we find that
And the first focus coordinate on the  x  axis is:
And the second focus coordinate on the  x  axis is:
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 14 − 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 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 (4x2 − 16x) + ( 9y2 + 108y) + 304 = 0 41x2 − 50cx + 25c2 − 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 − 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 15 − 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: