How do you identity if the equation #x^2+y^2+6y+13=40# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Jul 2, 2017

Answer:

It is a circle.

You graph it by:

  1. set your compass to a radius of #sqrt365/2#
  2. put the center at the point #(-13/2, -3)#
  3. draw the circle.

Explanation:

I suspect that the intended equation is:

#x^2+y^2+6y+13x=40" [1]"#

Otherwise, the 13 and 40 would have been combined into a single constant term.

The reference [Conic Section](https://en.wikipedia.org/wiki/Conic_section#General_Cartesian_form) tells us that, because the coefficient of the #x^2# term is equal to the coefficient of the #y^2# term, we know that the equation describes a #color(red)("circle")#. Here is a graph to prove it:

graph{x^2+y^2+6y+13x=40 [-30, 30, -15, 15]}

We can make fit the general Cartesian form for the equation of a circle:

#(x - h)^2 + (y-k)^2 = r^2" [2]#

where #(h,k)# is the center and r is the radius.

By completing the squares:

#x^2+13x+h^2+y^2+6y+k^2=40+h^2+k^2" [1.1]"#

We know, from their respective binomial expansions, that:

#-2hx = 13x# and #-2ky = 6y#

Solve for h and k:

#h=-13/2# and #k = -3#

We can obtain a value for r by substituting these values to the right side of equation [1.1}:

#r^2=40+(-13/2)^2+ (-3)^2#

#r^2 = 365/4#

#r = sqrt365/2#

Substituting these values into equation [2]:

#(x - (-13/2))^2 + (y-(-3))^2 = (sqrt365/2)^2" [2]"#

This is the standard Cartesian form with its center at #(-13/2,-3)# and a radius of #sqrt365/2#