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

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Answer 1 · 2016-11-22T06:23:12

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

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]}