How do you write the standard form of the hyperbola #9x^2-4y^2-90x+32y-163=0#?

1 Answer
Jun 1, 2018

Answer:

#((x-5)^2)/(6^2)-((y-4)^2)/(9^2)=1#

Explanation:

The standard form for a hyperbola (that opens sideways) is
#color(white)("XXX")((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1#

#9x^2-4y^2-90x+32y-163=0#

#rArr (9x^2-90x)-(4y^2-32y)=163#

#rArr color(green)9(x^2-10x)color(lime)(-4)(y^2-8y)=163#

#rArr color(green)9(x^2-10xcolor(blue)(+25))-color(lime)4(y^2-8ycolor(red)(+16))=163+color(green)9 * color(blue)(25) + (color(lime)(-4)) * 16#

#rArr color(green)9(x-5)^2-color(lime)4(y-4)^2=324#

#rArr (color(green)9(x-5)^2)/(324)-(color(lime)4(y-4)62)/324=1#

#rArr ((x-5)^2)/36-((y-4)^2)/81=1#

#rArr ((x-5)^2)/(6^2)-((y-4)^2)/(9^2)=1#