How do you find the standard form of #x^2 + 4y^2 - 8x + 16y + 16 =0# and what kind of a conic is it?

1 Answer
Dec 12, 2015

Answer:

An ellipse with center #(4,-2)#, x-axis radius: #4#, and y-axis radius: #2#

Explanation:

Given:
#color(white)("XXX")x^2+4y^2-8x+16y+16=0#

Group #x# and #y# terms (and constant) separately as
#color(white)("XXX")color(blue)(x^2-8x)+color(red)(4(y^2+4y))+color(green)(16)=0#

Complete the squares for both the #x# and the #y# sub-expressions:
#color(white)("XXX")color(blue)(x^2-8x+16) + color(red)(4(y^2+4y+4))+color(green)(16)-color(blue)(16)-color(red)(4(4))=0#

Reduce #x# and #y# sub-expressions to squared binomials and move constant to the right side:
#color(white)("XXX")(x-4)^2+4(y+2)^2= 16#
or
#color(white)("XXX")(x-4)^2+4(y+2)^2= 4^2#

Divide both sides by #(4^2)#
and replace #(y+2)# with #(y-(-2))#
#color(white)("XXX")((x-4)^2)/(4^2)+(y-(-2))^2/(2^2)=1#

Note that the general form of an ellipse is
#color(white)("XXX")(x-h)^2/(a^2)+((y-k)^2)/(b^2)=1#
with
#color(white)("XXX")#center at #(h,k)#,
#color(white)("XXX")#x-axis radius: #(a)#, and
#color(white)("XXX")#y-axis radius: #(b)#
graph{x^2+4y^2-8x+16y+16=0 [-2.195, 10.29, -4.895, 1.345]}