Conic Sections
Ellipse Equation of the ellipse, standard equation of the ellipse

The equation of the ellipse examples
Equation of the ellipse, standard equation of the ellipse
If in the direction of axes we introduce a coordinate system so that the center of the ellipse coincides with the
 origin, then coordinates of foci are F1(-c, 0) and F2( c, 0). For every point P(x, y) of the ellipse, according to definition  r1 + r2 = 2a, it follows that after squaring  and reducing Repeated squaring and grouping gives
(a2 - c2) · x2 + a2y2 = a2 · (a2 - c2),
 and since a2 - c2 = b2
 follows b2x2 + a2y2 = a2b2 equation of the ellipse, and after division by  a2b2, standard equation of the ellipse.
It follows from the equation that an ellipse is defined by values of a and b, or as they are associated through
the relation
a2 - c2 = b2,  we can say that it is defined by any pair of these three quantities.
 Intersections of an ellipse and the coordinate axes we determine from equation by putting,
y = 0  =>   x = + a, so obtained are vertices at the ends of the major axis A1(-a, 0) and A2(a, 0), and
x = 0  =>   y = + b, obtained are co-vertices, the endpoints of the minor axis B1(0, b) and B2(0, -b).
The line segments A1A2 = 2a and B1B2 = 2b are the major and minor axes while a and b are the
semi-major and semi-minor axes respectively. So the arc of the radius a centered at B1and B2 intersects the major axis at the foci F1and F2.
The focal parameter, called latus rectum and denoted 2p, is the chord perpendicular to the major axis
passing through any of the foci, as shows the above figure. The length of which equals the absolute value of ordinates of the points of the ellipse whose abscissas
x = c or x = -c that is so the length of the latus rectum the length of the semi-latus rectum of the ellipse.
The equation of the ellipse examples
Example:  Given is equation of the ellipse 9x2 + 25y2 = 225, find the lengths of semi-major and semi-minor axes, coordinates of the foci, the eccentricity and the length of the semi-latus rectum.
 Solution:  From the standard equation we can find the semi-axes lengths dividing the given
 equation by 225, coordinates of the foci F1(-c, 0) and F2( c, 0), since  Example:  From given quantities of an ellipse determine remaining unknown quantities and write equation of the ellipse, Solution:    a)  Using therefore, the semi-minor axis the linear eccentricity the semi latus rectum and the equation of the ellipse   the eccentricity and the equation of the ellipse  the semi latus rectum and the equation of the ellipse d) unknown quantities expressed through given values, Example:   Find the equation of the ellipse whose focus is F2(6, 0) and which passes through the point A(5Ö3, 4).
Solution:   Coordinates of the point A(5Ö3, 4) must satisfy equation of the ellipse, therefore thus, the equation of the ellipse Example:   Write equation of the ellipse passing through points A(-4, 2) and B(8, 1).
Solution:   Given points must satisfy equation of the ellipse, so Therefore, the equation of the ellipse or   x2 + 16y2 = 80.
Example:  In the ellipse 4x2 + 9y2 = 144 inscribed is a rectangle whose vertices lie on the ellipse and whose sides are parallel with ellipse axes. Longer side, which is parallel to the major axis, relates to the shorter side as 3 : 2. Find the area of the rectangle.
Solution:  It follows from the given condition that the coordinates of vertices of the rectangle must satisfy the
 same ratio, i.e.,      x : y = 3 : 2   =>   x = 3y/2. To determine points of the ellipse of which coordinates are in this ratio, put these variable coordinates into equation of the ellipse, P(3y/2, y)  =>   4x2 + 9y2 = 144 4(3y/2)2 + 9y2 = 144  =>    18y2 = 144, y1,2 = ±Ö144/18 = ±2Ö2,    x = 3y/2  =>  x1,2 = ±3Ö2. Therefore, the vertices of the rectangle, A(3Ö2, 2Ö2)B(-3Ö2, 2Ö2)C(-3Ö2, -2Ö2 and  D(3Ö2, -2Ö2).
The area of the rectangle A = 4 · x  · y = 4 · (3Ö2 ) · (2Ö2 ) = 48 square units.   Intermediate algebra contents 