16(x)^2 - (y)^2 - 160x - 12y + 360 = 0 The final answer is: ((y+6)^2)/4 - ((x-5)^2)/(1/4) = 1

Question

This is from a question asking for the conversion of the given equation into the standard form of the equation of a hyperbola! I'm having issues understanding a basic principle of getting the final answer.  I don't understand why the x-coordinate has the denominator of 1/4 in the final answer.

Answer 1

16x^2-y^2-160x-12y+360=0

(16x^2-160x)-(y^2+12y)=-360

16(x^2-10x)-(y^2+12y)=-360

add 400 to 16(x^2-10x) and add -36 to (y^2+12y)

16(x^2-10x+25)-(y^2+12y+36)=-360+400-36 (add 400-36 to the right side)

16(x-5)^2-(y+6)^2=4  after simplifying the right side

divide both sides by 4

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

Answer 2

Rearrange the terms to get:  (16x² - 160x) - (y² + 12y) = -360

 

16(x² - 10x) - (y² + 12y) = -360

 

Complete the square for the x-terms and for the y terms:

 

16(x² - 10x + 25) - (y² + 12y + 36) = -360 + 16(25) - 36

 

16(x-5)² - (y+6)² = 4

 

Divide by 4:  4(x-5)² - (y+6)²/4 = 1

 

4(x-5)² = (x-5)²/(1/4)

 

We have:  (x-5)²      (y+6)² 

                  -------  -    -------     = 1

                   (1/4)         4