(1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1 I was given this Ellipse in 'standard form' How can I find the Foci and Length of Major axis?
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Originally Posted by 14041471 (1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1 Compare the above to the formula they gave you, which probably looks like the following: . . . . .$\displaystyle \frac{(x\, -\, h)^2}{a^2}\, +\, \frac{(y\, -\, k)^2}{b^2}\, =\, 1$ How do h, k, a, and b relate to the foci and the axes? Have fun!
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