(x+4)^2/9-(y+3)^2/16=1 State the center, vertices,equation of asymptotes,and the foci.Graph the hyperbolas

Question

graphong hyperbolas

Answer 1

(x+4)^2/9-(y+3)^2/16=1

 

Since the x part is added,it is  horizontal  hyperbola

 

compare this equation with standard hyperbola(here horizontal) equation

(x-h)^2/a^2 -(y-k)^2/b^2=1

 

then a^2 = 9 and b^2 = 16,
so a = 3 and b =4.

Also, this hyperbola's foci and vertices are to the left and right of the center, on a horizontal line paralleling the x-axis.

 

From the equation, clearly the center is at (h, k) = (–4, -3).

Since the vertices are a = 3 units to either side, then they are at (-4-3, -3)=(-7,-3) and at (-4+3, -3)=(-1,-3).

The equation   c^2 =a^2 + b^2
gives                         c^2 = 9 + 16 = 25,
so                              c = 5,

And the foci, being 5 units to either side of the center, must be at (-4-5,-3)=(–9, -3) and (-4+5,-3)=(1, -3). 

 

Equation of asymptotes is 

 

y=± b/a (x-h) +k

 a is in the denominator of the slopes of the asymptotes, giving 

m = ± 4/3.

Keeping in mind that the asymptotes go through the center of the hyperbola,
The asymptotes are then given by the straight-line equations

 

y – (-3) = ± (4/3)(x -(-4))

 

y+3 =± 4/3  (x+4)

Center (–4, -3)

Vertices (-7,-3) and (-1,-3)

Foci (–9,-3) and (1, -3)

asymptotes equation 

y+3 = ±4/3  (x+4)