Question

Find an equation that models a hyperbolic lens with a = 16 inches and foci that are 40 inches apart. assume that the center of the hyperbola is the origin and the transverse axis is vertical.

Answer 1

took the test :-) it's C. y^2/256-x^2/144=1

Answer 2

The answer is:

`( y^(2) )/(256) } - ( x^(2) )/( 144 ) = 1`

Step-by-step explanation:
The general equation for a hyperbola with transverse axis vertical is

`( (y - k)^(2) )/( a^(2) ) - ( (x - h)^(2) )/( b^(2) ) = 1`
where:
h = x-coordinate of the center
k = y-coordinate of the center
a = semi-major axis
b = semi-minor axis

We know that the center has coordinates (0, 0), therefore:
h = 0
k = 0

We also know that
a = 16
2c = distance between foci = 40

In order to find b², we need to use the formula
c² = a² + b²

and solve it for b²:
b² = c² - a²
= (40÷2)² - 16²
= 400 - 256
= 144

Now we can substitute the obtained values in the general equation to find the required equation:

`( y^(2) )/(256) } - ( x^(2) )/( 144 ) = 1`