How do you write the standard form of the hyperbola #-2x^2+3y^2+4x-60y+268=0#?

2 Answers
Oct 17, 2016

Answer:

#-(x-1)^2/(sqrt15)^2+(y-10)^2/(sqrt10)^2=1#

Explanation:

#-2x^2+3y^2+4x-60y+268=0#
Rearranging
#-2(x^2-2x)+3(y^2-20y)=-268#
Completing the squares
#-2(x^2-2x+1)+3(y^2-20y+100)=-268-2+300#
#-2(x-1)^2+3(y-10)^2=30#
Dividing by 30
#-(x-1)^2/15+(y-10)^2/10=1#
In standard form
#-(x-1)^2/(sqrt15)^2+(y-10)^2/(sqrt10)^2=1#

Oct 17, 2016

Answer:

Please see the explanation for the process.

#(y - 10)^2/(sqrt(10))^2 -(x - 1)^2/(sqrt(15))^2 = 1#

Explanation:

Add #3k^2 - 2h^2 - 268# to both sides:

#-2x^2 + 4x - 2h^2 + 3y^2 - 60y + 3k^2 = 3k^2 - 2h^2 - 268#

Remove a factor of -2 from the first 3 terms and a factor of 3 from the next 3 terms:

#-2(x^2 - 2x + h^2) + 3(y^2 - 20y + k^2) = 3k^2 - 2h^2 - 268#

Using the pattern, #(x - h)^2 = x^2 -2hx + h^2#, set the right side of the pattern equal to the first three terms:

#x^2 -2hx + h^2 = x^2 - 2x + h^2#

Combine like terms:

#-2hx = - 2x#

Solve for #h# and compute #h^2#:

#h = 1, h^2 = 1#

Substitute #(x - 1)^2# for #x^2 - 2x + h^2# and #-2# for #-2h^2#

#-2(x - 1)^2 + 3(y^2 - 20y + k^2) = 3k^2 - 2 - 268#

Using the pattern, #(y - k)^2 = y^2 -2ky + k^2#, set the right side of the pattern equal to the second three terms:

#y^2 -2ky + k^2 = y^2 - 20y + k^2#

Combine like terms:

#-2ky = -20y#

#k = 10, k^2 = 100#

Substitute #(y - 10)^2# for #y^2 - 20y + k^2# and #300# for #3k^2#

#-2(x - 1)^2 + 3(y - 10)^2 = 300 - 2 - 268#

Combine the terms on the right:

#-2(x - 1)^2 + 3(y - 10)^2 = 30#

Divide both sides by 30:

#(y - 10)^2/10 -(x - 1)^2/15 = 1#

Write the denominators as squares:

#(y - 10)^2/(sqrt(10))^2 -(x - 1)^2/(sqrt(15))^2 = 1#