How do you find the standard form of #4x^2 - 5y^2 - 16x - 30y - 9 = 0#?

1 Answer
Dec 17, 2015

Answer:

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

Explanation:

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

Given
#color(white)("XXX")4x^2-5y-16x-30y-9=0#

Group the #x# terms and the #y# terms separately and transfer the constant to the right side of the equation.
#color(white)("XXX")color(red)(4x^2-16x)-color(blue)(5y^2+30y) = color(green)(9)#

Extract the constant factors from each of the #x# and #y# sub-expressions.
#color(white)("XXX")color(red)(4(x^2-4x))-color(blue)(5(y^2+6y)) = color(green)(9)#

Complete the squares for each of the #x# and #y# sub-expressions
#color(white)("XXX")color(red)(4(x^2-4x+2^2))-color(blue)(5(y^2+6y+3^2)) = color(green)(9)+color(red)(16)-color(blue)(45)#

#color(white)("XXX")color(red)(4(x-2)^2)-color(blue)(5(y+3)^2) = -20#

Dividing both sides by #(-20)#
#color(white)("XXX")color(red)(-(x-2)^2/5)color(blue)(+(y+3)^2/4) = 1#

Convert the denominators into squares and re-order the left side terms to avoid the ugly leading minus sign.
#color(white)("XXX")color(blue)((y+3)^2/2^2)-color(red)((x-2)^2/(sqrt(5))^2) = 1#

Note that the reversal of the #x# and #y# terms from the normal form implies that the transverse axis is vertical.
graph{4x^2-5y^2-16x-30y-9=0 [-9.14, 19.34, -9.06, 5.18]}