Question

Foci at \( (0,-12) \) and \( (0,12) \); asymptote the line \( y=-x \)

Answer 1

Explanation:

Since the foci lies on the y axis, we will be using the vertical ellipse equation

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

Notice that the foci is (0,-12) and (0,12), that means the center,(h,k) is (0,0)

So our new equation is

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

To find the asymptotes of the hyperbola , set it equal to

`\frac{ {y}^(2) }{ {a}^(2) } - \frac{ {x}^(2) }{ {b}^(2) } = 0`

Difference of Squares

`( (y)/(a) + (x)/(b) )( (y)/(a) - (x)/(b) ) = 0`

Set both equations equal to zero , and solve for y

`y = - (a)/(b) x`

or

`y = (a)/(b) x`

Notice that one of our asymptotes slope is -1. That means a and b are both the same number, so we can represent them as a singular variable,u,.

The equation of a focus is

`c = \sqrt{ {a}^(2) + {b}^(2) }`

Here, c is 12

`12 = \sqrt{ {a}^(2) + {b}^(2) }`

Remember that both a and b are equal to each other so let u =a, and u=b

`12 = \sqrt{ {u}^(2) + {u}^(2) }`

`12 = \sqrt{2 {u}^(2) }`

`144 = 2 {u}^(2)`

`72 = {u}^(2)`

`u = 6 √(2)`

So our equation is

`\frac{ {y}^(2) }{72} - \frac{ {x}^(2) }{72} = 1`