If I wanted to find the vertices of the ellipse x^2/4 + y^2/25 = 1, what would I have to do?

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- Mar 31st 2007, 03:12 PMamanda1Vertices ellipse
If I wanted to find the vertices of the ellipse x^2/4 + y^2/25 = 1, what would I have to do?

- Mar 31st 2007, 04:13 PMPlato
The principal axis of this ellipse is vertical. WHY?

The vertices are then on that axis.

What are the maximum y-values? - Mar 31st 2007, 04:24 PMJhevon
the eqution of an ellipse with foci (0,+/- c) and vertices (0,+/-a) and center the origin (0,0), is given by the equation:

x^2/b^2 + y^2/a^2 = 1

where c^2 = a^2 - b^2

note: a is always bigger than b. if the number under the x was bigger, we would call that a^2.

the vertices of an ellipse lie along the line connecting its two foci. now there are technical ways to know whether the foci lies along the x-axis or y-axis for an ellipse with center the origin, but the easiest way is to see which denominator is bigger. is the number dividing x^2 bigger, or is the number dividing y^2 bigger? if the number under x^2 is bigger, then the foci (and hence the vertices) will be on the x-axis, and vice versa is the number under the y^2 is bigger.

so in the equation: x^2/4 + y^2/25 = 1, since the number under the y^2 is bigger, this number is termed a^2. so a = sqrt(25) = 5. this is the bigger denominator, so the vertices will be vertical along the y-axis in this case, so the vertices are given by (0,+/-a), which means**the vertices of this ellipse is (0,+/-5)**...notice that this is, as Plato says, a vertical line, it runs along the y-axis