Question

Identify the standard form of the equation by completing the square.

4x2 − 9y2 − 8x + 36y − 68 = 0

Answer 1

`((x-1)^2)/(9)-((y-2)^2)/(4)=1`

Explanation:

Given equation:

`4x^2-9y^2-8x+36y-68=0`

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

`\implies 4x^2-8x-9y^2+36y=68`

Factor out the coefficient of the x² term and the y² term.

`\implies 4(x^2-2x)-9(y^2-4y)=68`

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

`\implies 4\left(x^2-2x+\left((-2)/(2)\right)^2\right)-9\left(y^2-4y+\left((-4)/(2)\right)^2\right)=68+4\left((-2)/(2)\right)^2-9\left((-4)/(2)\right)^2`

`\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36`

Factor the two perfect trinomials on the left side:

`\implies 4(x-1)^2-9(y-2)^2=36`

Divide both sides by the number of the right side so the right side equals 1:

`\implies (4(x-1)^2)/(36)-(9(y-2)^2)/(36)=(36)/(36)`

Simplify:

`\implies ((x-1)^2)/(9)-((y-2)^2)/(4)=1`

Therefore, this is the standard equation for a horizontal hyperbola with:

• center = (1, 2)
• vertices = (-2, 2) and (4, 2)
• co-vertices = (1, 0) and (1, 4)
• 
  `\textsf{Asymptotes}: \quad y = -(2)/(3)x+(8)/(3) \textsf{ and }y=(2)/(3)x+(4)/(3)`
• 
  `\textsf{Foci}: \quad (1-√(13), 2) \textsf{ and }(1+√(13), 2)`