How do you write the equation#9x^2 - y^2 - 18x + 6y - 36 = 0# in standard form and what type of conic is it?

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Answer 1 · 2017-06-28T06:59:16

It is a hyperbola.

#9x^2-y^2-18x+6y-36=0# can be written as

#(9x^2-18x)-(y^2-6y)-36=0#

or #9(x^2-2x+1)-(y^2-6y+9)-36-9+9=0#

or #9(x-1)^2-(y-3)^2=36#

or #(x-1)^2/4-(y-3)^2/36=1#

As it is of the type #(x-h)^2/a^2-(y-k)^2/b^2=1#

it is a hyperbola with center at #(1,3)#

It is a hyperbola.

The graph appears as follows:

graph{9x^2-y^2-18x+6y-36=0 [-20, 20, -6.88, 13.12]}