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

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Answer 1 · 2018-03-30T13:55:29

#(x+4)^2/(sqrt12)^2+(y-1)^2/(sqrt8)^2=1#

#2(x+4)^2+3(y-1)^2=24# - dividing by #24#, it can be written as

#(2(x+4)^2)/24+(3(y-1)^2)/24=1#

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

or #(x+4)^2/(sqrt12)^2+(y-1)^2/(sqrt8)^2=1#

which is the equation of an ellipse with center at #(-4,1)#,

major axis parallel to #x#-axis is #2sqrt12=4sqrt3#

and minor axis parallel to #y#-axis is #2sqrt8=4sqrt2#

graph{((x+4)^2/12+(y-1)^2/8-1)((x+4)^2+(y-1)^2-0.01)=0 [-11, 3, -2.5, 4.5]}