Given: #4x^2-5y^2-40x-20y+160=0#
Subtract 160 from both sides:
#4x^2-5y^2-40x-20y=-160#
Add #4h^2# to both sides:
#4x^2-40x+4h^2-5y^2-20y=-160+4h^2#
Remove the common factor, 4, from the first 3 terms:
#4(x^2-10x+h^2)-5y^2-20y=-160+4h^2#
Subtract #5k^2# from both sides:
#4(x^2-10x+h^2)-5y^2-20y-5k^2=-160+4h^2-5k^2#
Remove a common factor of -5 from the next 3 terms:
#4(x^2-10x+h^2)-5(y^2+4y+k^2)=-160+4h^2-5k^2" [1]"#
Because we are looking for the square #(x-h)^2 = x^2-2hx+h^2# we use the equation:
#-2hx = -10x#
To find the value of h:
#h = 5#
Into equation [1], substitute #(x-5)^2" for " x^2-10x+h^2# and 100 for #4h^2#:
#4(x-5)^2-5(y^2+4y+k^2)=-160+100-5k^2" [1.1]"#
Because we are looking for the square #(y-k)^2 = y^2-2ky+k^2# we use the equation:
#-2ky = 4y#
To find the value of k:
#k = -2#
Into equation [1.1], substitute #(y-(-2))^2" for " y^2+4y+k^2# and -20 for #-5k^2#:
#4(x-5)^2-5(y-(-2))^2=-160+100-20" [1.2]"#
Combine like terms on the right:
#4(x-5)^2-5(y-(-2))^2=-80" [1.3]"#
Divide both sides by -80:
#(y-(-2))^2/16-(x-5)^2/20=1" [1.3]"#
Write the denominators as squares:
#(y-(-2))^2/(4)^2-(x-5)^2/(2sqrt5)^2=1" [1.4]"#
Equation [1.4] is standard form.
