25 4x
25 4x
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How do you find the vertices and foci of 4x^2 + 25y^2 = 25?
Well, what I'm mainly asking is how do you simplify it to a form where you can find the foci and vertices.... i've tried but can't figure it out.
Thanks!
First, you want to get the relation in the form:
(x/a)^2 + (y/b)^2 = 1
where a and b are the semi-major and semi-minor axes, with the larger of a and b representing the major axis and the lesser, the minor.
Because the constant at the end is 25, divide by that number.
4x^2 + 25y^2 = 25
4x^2/25 + y^2 = 1
(2x/5)^2 + (y/1)^2 = 1
So, in the above formula, a = 5/2 and b = 1. As a>b, the major axis is horizontal and a is the semi-major axis, that is, the length from the centre to the edge of the ellipse, or half the length of the major axis; b is the semi-minor axis.
The vertices of the major axis are (-a, 0), (a, 0). The vertices of the minor axis are (0, -b), (0, -b). The foci are (-c, 0), (c, 0), where c=sqrt(a^2-b^2). The eccentricity of the ellipse is e=sqrt(1-(b/a)^2).
In this ellipse, the major axis vertices are (-5/2, 0) and (5/2, 0). The minor axis vertices are (0, -1) and (0, 1). For the foci, c = sqrt(25/4 - 1) = sqrt(21/4) = sqrt(21)/2. So the foci are (-sqrt(21)/2, 0) and (sqrt(21)/2, 0).