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

1 Answer
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