How do you graph #4x^2+9y^2+24-18+9=0#?

1 Answer
Sep 8, 2016

Answer:

(see below)

Explanation:

Note: I am assuming that the equation should have been:
#color(white)("XXX")4x^2+9y^2+24color(red)(x)-18color(red)(y)+9=0#

The standard form of an ellipse is
#color(white)("XXX")((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1#

with center at #(h,k)#,
radius on X-axis #=a#, and
radius on Y-axis#=b#

Converting the (assumed) equation into this form:
#color(white)("XXX")4(x^2+6x)+9(y^2-2y)=-9#

#color(white)("XXX")4(x^2+6x+9)+9(y^2-2y+1)=-9+4xx9+9xx1#

#color(white)("XXX")4(x+3)^2+9(y-1)^2=36#

#color(white)("XXX")((x+3)^2)/(3^2)+((y-1)^2)/(2^2)=1#

So we need to draw an ellipse with center at #(-3,1)#,
a radius of #3# parallel to the X-axis, and
a radius of #2# parallel to the Y-axis.

enter image source here