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

Question

2207 views around the world

You can reuse this answer
Creative Commons License

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

enter image source here

How do you find the standard form of 5x^2 + 8y^2 + 16y - 32 = 05x^2 + 8y^2 + 16y - 32 = 0 and what kind of a conic is it?