How do you identity if the equation #y^2+18y-2x=-84# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Apr 1, 2017

Answer:

It is a parabola.

Explanation:

An equation which can be written in the form of #y=ax^2+bx+c# or #x=py^2+qy+r# is a parabola and to draw it you have to just convert it into vertex form such as #y=a(x-h)^2+k# or #x=p(y-k)^2+h#. Here #(h,k)# is the vertex.

Here we have #y^2+18y-2x=-84# or #2x=y^2+18y+84# and in vertex form it is

#x=1/2(y^2+18y)+42#

or #x=1/2(y^2+2xx9y+9^2-9^2)+42#

or #x=1/2(y+9)^2-81/2+42#

or #x=1/2(y-(-9))^2+3/2#

Here #y+9=0# is the axis of symmetry and as coefficient of #(y+9)^2# is #1/2#, a positive number, it opens up towards right. Its vertex is #(3/2,-9)#.

We can now choose some values of #y# around #-9#, say #{-6,-7,-8,-10,-11,-12}# and find corresponding #x# which are #{6,7/2,2,2,7/2,6}#.

This gives us points #(6,-6)#, #(7/2,-7)#, #(2,-8)#, #(3/2,-9)#, #(2,-10)#, #(7/2,-11)#, #(6,-12)# and joining them gives us the following graph.

graph{y^2+18y-2x=-84 [-10.5, 29.5, -18.64, 1.36]}