How do you find the standard form of #5x^2 + 8y^2 + 16y - 32 = 0# and what kind of a conic is it?

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Answer 1 · 2016-01-18T20:43:31

First, complete the square ...

#5x^2 + 8y^2 + 16y - 32 = 0#

#5x^2 + 8(y^2 + 2y) = 32#

Now, complete the square ...

#5x^2 + 8(y^2 + 2y+1) = 32+8#

#5x^2 + 8(y+1)^2 = 40#

Now, divide both sides by 40 ...

#x^2/8+(y+1)^2/5=1#

This is an ellipse since the coefficients on x and y are different.

The center #=(0,-1)#.

The major axis is horizontal .

Hope that helped

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