Question

How do you find the standard form of `2y^2+9x^2=9-x` and what kind of a conic is it?

Answer 1

This is an ellipse, `(x+1/18)^2/(2917/2916)+y^2/(2917/648)=1`

Explanation:

Let's rewrite the equation

`9x^2+2y^2+x-9=0`

Compare this to the general equation of the conics

`Ax^2+Bxy+Cy^2+Dx+Ey+F=0`

Let's calculate the dscriminant,

`Delta=B^2-4AC=0-4*9*2=-72`

As, #Delta<0, we expect an ellipse

Completing the squares

`9(x^2+x/9)+2y^2=9`

`9(x^2+x/9+(1/18)^2)+2y^2=9+(1/18)^2`

`9(x+1/18)^2+2y^2=2917/324`

`(9(x+1/18)^2)/(2917/324)+(2y^2)/(2917/324)=1`

`(x+1/18)^2/(2917/2916)+y^2/(2917/648)=1`

This is the standard equation of the ellipse

the center is `(-1/18,0)`

graph{9x^2+x+2y^2-9=0 [-4.382, 4.386, -2.19, 2.192]}