Graph x^2-3y^2+8x-6y+4=0

Problem

x2−3*y2+8*x−6*y+4=0

Solution

1. Group the x terms and y terms together and move the constant to the right side of the equation.

(x2+8*x)−3*(y2+2*y)=−4

2. Complete the square for both the x and y expressions. For x add (8/2)2=16 For y add (2/2)2=1 Remember to balance the equation by adding the same values to the right side, noting the coefficient −3 for the y term.

(x2+8*x+16)−3*(y2+2*y+1)=−4+16−3*(1)

3. Simplify the equation into its squared binomial forms and combine the constants on the right side.

(x+4)2−3*(y+1)2=9

4. Divide both sides by 9 to put the equation into the standard form of a hyperbola.

((x+4)2)/9−((y+1)2)/3=1

5. Identify the properties of the hyperbola. The center (h,k) is (−4,−1) Since the x term is positive, it is a horizontal hyperbola. The values are a2=9 (so a=3 and b2=3 (so b=√(,3).

6. Determine the vertices and asymptotes. The vertices are at (−4±3,−1) which are (−7,−1) and (−1,−1) The slopes of the asymptotes are ±b/a=±√(,3)/3

Final Answer

x2−3*y2+8*x−6*y+4=0⇒((x+4)2)/9−((y+1)2)/3=1

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