Question

The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.

Answer 1

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = $\frac{b}{a}$

- This is a vertical hyperbola, the equation is

       $\frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1$

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

       $\frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1$

       $\frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1$

Answer 2

Answer:

$\frac{(x+4)^{2}}{4} - \frac{(y-3)^{2}}{16} = 1$

Step-by-step explanation:

The hyperbola centered at (h,k) has the following expression:

$\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$

Where $a$ and $b$ are the length of the horizontal and vertical semi-axes, respectively.

Since the center and one vertex share the same vertical component ($x=-4$), it is easy to conclude that hyperbola has a vertical configuration ($b > a$). The distance between the center and the known vertex is equal to the length of the vertical semi-axis. Therefore:

$b = 4$

The slope of the hyperbola is given by the following relationship:

$\frac{b}{a} = 2$

The length of the horizontal semi-axis is:

$a = \frac{b}{2}$

$a = 2$

The standard form of the equation of the hyperbola is:

$\frac{(x+4)^{2}}{4} - \frac{(y-3)^{2}}{16} = 1$