Question

Write the standard form of the equation of the hyperbola 4x2 − 9y2 − 16x − 36y − 56 = 0.

Answer 1

We have that
4x² − 9y² − 16x − 36y − 56 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(4x² -16x) +(− 9y²− 36y)= 56

(4x² -16x) -(9y²+36y)= 56

Factor the leading coefficient of each expression

4(x²-4x) -9(y²+4y)= 56

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

4(x²-4x+4) -9(y²+4y+4)= 56+16-36

4(x²-4x+4) -9(y²+4y+4)= 36

Rewrite as perfect squares

4(x-2)² -9(y+2)²= 36

divide by 36 both sides

$[(1/9)( x-2)^{2}]-[(1/4)( y+2)^{2}]=1$

the answer is

The equation of the hyperbola in the standard form is

$[(1/9)( x-2)^{2}]-[(1/4)( y+2)^{2}]=1$