Question

Complete the standard form of the equation of a hyperbola that has vertices at (-10, -15) and (70, -15) and one of its foci at (-11, -15).

Answer 1

Answer:

The above answer is correct but the 3 should be a 9

Step-by-step explanation:

Plato

Answer 2

Answer:

$\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1$

Step-by-step explanation:

The equation of the horizontal hyperbola in standard form is:

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

The position of its center is:

$C(x,y) = \left(\frac{-10 + 70}{2}, -15 \right)$

$C(x, y) = (30,-15)$

The values for c and a are respectively:

$a = 70 - 30$

$a = 40$

$c = 30 - (-11)$

$c = 41$

The remaining variable is computed from the following Pythagorean identity:

$c ^{2} = a^{2} + b^{2}$

$b = \sqrt{c^{2}-a^{2}}$

$b = \sqrt{41^{2}-40^{2}}$

$b = 3$

Now, the equation of the hyperbola is:

$\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1$