How do you identity if the equation #x^2+2y^2=2x+8# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Apr 9, 2017

Answer:

#x^2+2y^2-2x-8=0# is an ellipse.

Explanation:

Let the equation be of the type #Ax^2+Bxy+Cy^2+Dx+Ey+F=0#

then if

#B^2-4AC=0# and #A=0# or #C=0#, it is a parabola

#B^2-4AC<0# and #A=C#, it is a circle

#B^2-4AC<0# and #A!=C#, it is an ellipse

#B^2-4AC>0#, it is a hyperbola

This equation is #x^2+2y^2-2x-8=0# and hence, #A=1#, #B=0#, #C=2# and #B^2-4AC=-8#

As #B^2-4AC<0# and #A!=C#, hence it is an ellipse. We can write it as

#x^2-2x-8+2y^2=0#

or #x^2-2x+1+2y^2=9#

or #(x-1)^2/9+y^2/(9/2)=1#

or #(x-1)^2/3^2+y^2/(3/sqrt2)^2=1#

Hence, this is n ellipse with center at #(1,0)#, major axis along #x#-axis as #3xx2=6# and minor axis along #y#-axis #2xx3/sqrt2=3sqrt2#. Hence end points for major axis on #x#-axis are #(-2,0)# and #(4,0)# and end points for minor axis are #(1,3/sqrt2)# and #(1,-3/sqrt2)#

graph{y(x^2-2x-8+2y^2)(x-1)=0 [-4.21, 5.79, -2.46, 2.54]}