How do you write #4x^2 + y^2 – 4y – 12 = 0# in standard form and why type of conic is it?

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Answer 1 · 2016-02-18T07:33:20

#(x-0)^2/4+(y-2)^2/16=1#
Ellipse

by completing the square
#4x^2+y^2-4y-12=0#

#4x^2+y^2-4y+4-4-12=0#

#4(x-0)^2+(y-2)^2=16#

#(4(x-0)^2)/16+(y-2)^2/16=16/16#

#(x-0)^2/4+(y-2)^2/16=1#

Kindly see the graph

graph{(x-0)^2/4+(y-2)^2/16=1[-20,20,-10,10]}

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