How do you identity if the equation #5x^2+6x-4y=x^2-y^2-2x# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Jan 11, 2017

Answer:

It is an ellipse.

Explanation:

Let us write the equation in the form #Ax^2+Bxy+Cy^2+Dx+Ey+F=0#

As #5x^2+6x-4y=x^2-y^2-2x# is equivalent to

#5x^2-x^2+y^2+6x+2x-4y=0#

or #4x^2+y^2+8x-4y=0#

Hence, here we have #A=4#, #B=0# and #C=1#

and therefore #B^2-4AC=0-4xx4xx1=-16 < 0# and #A!=C#

Hence it is an ellipse (for details see here).

The equation can be written as #4(x^2+2x+1)+(y^2-4y+4)=8#

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

or #(x+1)^2/2+(y-2)^2/8=1# and comparing it with general form of equation of an ellipse #(x-h)^2/a^2+(y-k)^2/b^2=1#, where center is #(h,k)# and axes of symmetry are #x=h# and #y=k#. The two axes are #2a# and #2b#.

Here, we have center at #(-1,2)# and axes of symmetry are #x=-1# and #y=2#. It has major axis #2sqrt8=4sqrt2# (vertically aligned as #b>a#) and minor axis is #2sqrt2#.

graph{5x^2+6x-4y=x^2-y^2-2x [-11.75, 8.25, -2.84, 7.16]}