What kind of conic is defined by the equation #2x^2+4y^2-4x+12y=0#?

1 Answer
Manasvini Nittala
Oct 24, 2017

the conic is an ellipse with centre #(1,-3/2)# and eccentricity #1/sqrt2#

Explanation:

the given equation is #2x^2+4y^2-4x+12y=0#
we are basically trying to convert the equation in form of perfect squares.
on doing so the equation can be written as #2(x^2-2x+1)+4(y^2+3y+(3/2)^2)=2+9#
#rArr2(x-1)^2+4(y+3/2)^2=11#
#rArr(2(x-1)^2)/11+(4(y+3/2)^2)/11=1#
#rArr(x-1)^2/(11/2)+(y+3/2)^2/(11/4)=1#
this is of the form #((x-h)^2)/a^2+((y-k)^2)/b^2=1#
which is the general form of an elipse
hence centre is #(h,k)=(1,-3/2)# and eccentricity #e=sqrt((a^2-b^2)/a^2)=1/sqrt2#

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