Algebra 2 Tutorial
Conic sections : vertices, foci and eccentricity of an ellipse.
From the equation of a conic section, here an ellipse described by 9x^2 + 36y^2 = 324, we will find vertices, foci and eccentricity. Given the equation, an oval of center (H,K) can be traced. A path of points traced (x,y) so that the sum of its distances from two fixed points is constant. The two fixed points are called foci (plural noun of focus) and they are points of equal distance from the center.
*Side note : Contrary to a circle, there is no radius r in an ellipse because the distance from the center to its perimeter changes and creates two axis : the major axis (wide) and minor axis (narrow). The minor and major axis can be vertical, horizontal or even on an angle.
Vertices are points also equidistant from the centre, situated at the edges of the major axis. We will find the vertices to find the foci.
Given the equation 9x^2 + 36y^2 = 324 with points (x,y). We need to find the centre to find the vertices, foci and eccentricity.
The basic equation of an ellipse,
((x – H)^2 )/a^2 + ((x – K)^2 )/b^2 = 1 (here with a horizontal major axis, as a is under H)
Center = (H,K)
a = distance from center to an end of major axis (called a vertex; vertices)
b = distance from center to ends of minor axis
c = distance from center to a focus (same distance for the two foci)
Foci : (H+c,K) and (H-c,K)
The graph of an ellipse gives a^2 = b^2 + c^2
c^2 = a^2 – b^2 (NOT the Pythagorean theorem)
eccentricity : e = c/a
Extra info: (Lenght of) Major axis = 2a (Lenght of) Minor axis = 2b
Simplify 9x^2 + 36y^2 = 324 by dividing each side by 324 to get
(x^2)/6^2 + (y^2)/3^2 = 1
a = 6 and b = 3
Centre (H,K) : (0,0)
Vertices : Major(-6,0);(6,0) and minor (-3,0);(3,0)
c^2 = a^2 – b^2
c = 3(3^1/2) or 3 times the square root of 3
Foci : (3(3^1/2),0) and (-3(3^1/2),0)
e = c/a
e = 3(3^1/2)/6 = (3^1/2)/2
The ellipse 9x^2 + 36y^2 = 324 has vertices Major(-6,0);(6,0) and minor (-3,0);(3,0), foci (3(3^1/2),0) and (-3(3^1/2),0) and eccentricity (3^1/2)/2.
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