How do you find the standard form of #9x^2 + 4y^2 - 18x + 16y = 11# and what kind of a conic is it?

1 Answer
Dec 5, 2015

Answer:

standard form: #(x-1)^2/(2^2)+(y+2)^2/(3^2)=1#
which is an ellipse with center #(1,-2)#, radius along the x-axis of #2#, and radius along the y-axis of #3#

Explanation:

Given #9x^2+4y^2-18x+16y=11#

By regrouping and completing the squares:
#color(white)("XXX")9(x^2-2x+1)-9 + 4(y^2+4y+4)-16 = 11#

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

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

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

which is the standard form for an ellipse with the values given in the Answer above.

graph{9x^2+4y^2-18x+16y=11 [-6.22, 6.27, -5.113, 1.127]}