How do you write #2(x+4)^2 + 3(y-1)^2 = 24# in standard form?

1 Answer
May 3, 2016

Answer:

#(x-(-4))^2/(sqrt(12))^2 +(y-1)^2/(sqrt(8))^2=1#

Explanation:

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

Given
#color(white)("XXX")2(x+4)^2+3(y-1)^2=24#

#rArr#
#color(white)("XXX")(x+4)^2/12+(y-1)^2/8=1#

#color(white)("XXX")(x-(-4))^2/(sqrt(12))^2+(y-1)^2/(sqrt(8))^2=1#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If you like you could replace #(sqrt(12))^2# with #(2sqrt(3))^2#
and #(sqrt(8))^2# with #(2sqrt(2))^2#
...but this really doesn't make the form any simpler.