Question

9x^2+4y^2+54x+8y-59=0

Answer 1

The standard form of an ellipse is

$\dfrac{(x-x_0)^2}{a^2}+\dfrac{(y-y_0)^2}{b^2}=1$

Let's try to complete some squares: if we focus on the part involing x, we have

$9x^2+54x = 9x^2+54x+81-81 = (3x+9)^2-81 = 9(x+3)^2-81$

Similarly, for the part involving y, we have

$4y^2+8y = 4y^2+8y+4-4 = 4(y+1)^2-4$

So, the equation becomes

$9(x+3)^2-81 + 4(y+1)^2-4 - 59=0 \iff 9(x+3)^2+ 4(y+1)^2-81-4-59=0 \iff 9(x+3)^2+ 4(y+1)^2 = 144$

Divide both sides by 144 to get

$\dfrac{(x+3)^2}{16}+ \dfrac{(y+1)^2}{36} = 1$

which is the standard form