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

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

Explanation:

Plato

Answer 2

`((x-30)^(2))/(40^(2)) - ((y+15)^(2))/(3^(2)) = 1`

Explanation:

The equation of the horizontal hyperbola in standard form is:

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

The position of its center is:

`C(x,y) = \left((-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:

`((x-30)^(2))/(40^(2)) - ((y+15)^(2))/(3^(2)) = 1`