How to find the center, transverse axis, vertices, foci, and asymptotes of the following equation? 2X^2 - y^2 +4x + 4y -4 = 0

Question

This is a conics question for my college algebra class.

Answer 1

We complete the square to turn this equation into a factored form:

 

2(x^2+2x+2)-(y^2-4y+4)-4=0

2(x+sqrt(2))^2-(y-2)^2=4

[(x+sqrt(2))^2]/2-[(y-2)^2]/4=1 

This is an equation of a hyperbola with the form [(x-h)^2]/a^2-[y-k)^2]/b^2=1.

So in our case h=-sqrt(2), a=sqrt(2), k=2, b=2.

 

The center is at (h,k)= (-sqrt(2), 2). Since the negative is in front of the y term, the hyperbola's foci and vertices are to the left and right of the center. Also the transverse axis is parallel to the x axis through the center, vertices and foci. So the transverse axis is at y=2.

 

 

The  vertices are a=sqrt(2) units to the left and right so they are at (-2sqrt(2), 2) and (0, 2). 

 

The equation c^2 – a^2 = b^2 gives me c^2 = 2+ 4 = 6, so c =sqrt(6), and the foci, being sqrt(6) units to the left and right of the center, must be at (–sqrt(6)-sqrt(2), 2) and (sqrt(6)-sqrt(2), 2).

 

Since there was a positive in front of the x part of the equation, then a is in the denominator of the slopes of the asymptotes, giving me m = ± b/a= ±2/sqrt(2)= ±sqrt(2). Keeping in mind that the asymptotes go through the center of the hyperbola, the asymptotes are then given by the equations y = ± sqrt(2)(x +sqrt(2))+2.