Write the equation of the conic #18y^2-5x^2+72y+30x-63=0# in standard form?

1 Answer
Oct 25, 2017

#(y+2)^2/(sqrt2)^2-(x-3)^2/(6/sqrt5)^2=1#

Explanation:

This is the equation of typical hyperbola whose conjugate axis is parallel to #y#-axis. Its standard form is

#(y-k)^2/a^2-(x-h)^2/b^2=1# - whose center is #(h,k)#, major axis is #y=k# and conjugate axis is #x=h#.

We convert it into standard form as follows:

#18y^2-5x^2+72y+30x-63=0#

or #18(y^2+4y)-5(x^2-6x)=63#

or #18(y^2+4y+4)-5(x^2-6x+9)-72+45=63#

or #18(y+2)^2-5(x-3)^2=36#

or #(y+2)^2/2-(x-3)^2/(36/5)=1#

or #(y+2)^2/(sqrt2)^2-(x-3)^2/(6/sqrt5)^2=1#

and hence center is #(3,-2)#, major axis is #y=-2# and conjugate axis is #x=3#. ts graph appears as shown below:

graph{(18y^2-5x^2+72y+30x-63)((x-3)^2+(y+2)^2-0.03)(y+2)(x-3)=0 [-18.33, 21.67, -12.32, 7.68]}