Answer:
[tex]\frac{x^2}{25} + \frac{y^2}{16} = 1[/tex]
Step-by-step explanation:
Great, so the question already tells you that you're using the standard form of an ellipse. All you gotta do is apply your knowledge of the characteristics of an ellipse graph.
The standard form of an ellipse is:
[tex]1 = \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}[/tex]
[tex]h[/tex] and [tex]k[/tex] are the offset, or the coordinates of the center of the ellipse. The center of the ellipse will always be on (h,k). Now because the question says the center is (0,0), we can make h and k equal to 0 in the equation giving:
[tex]1 = \frac{x^2}{a^2} + \frac{y^2}{b^2}[/tex].
Now, the height of the ellipse will always be equal to 2a. The width will be equal to 2b.
Since we are told what the height and width should be, we can find the a and b values quite easily using algebra.
So first, height:
[tex]10 = 2a\\5 = a[/tex]
Now width:
[tex]8 = 2b\\4 = b[/tex]
Subbing for a and b in the equation give you:
[tex]1 = \frac{x^2}{5^2} + \frac{y^2}{4^2}\\=\\1 = \frac{x^2}{25} + \frac{y^2}{16}[/tex]
