Question

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

An ellipse in that needs to be in standard form
Please explain how to do

Answer 1

The standard form of an ellipse is

`((x-x_0)^2)/(a^2)+((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

`((x+3)^2)/(16)+ ((y+1)^2)/(36) = 1`

which is the standard form

Answer 2

D

edge

Explanation: