Acamera is placed in front of a hyperbolic mirror. the equation of the hyperbola that models the mirror is , where x and y are in inches. the camera is pointed toward the vertex of the hyperbolic mirror and is positioned such that the lens is at the nearest focus to that vertex.

The lens is 1 inch from the mirror

Step-by-step explanation:

* Lets revise the equation of the hyperbola with center (0 , 0) and

 transverse axis parallel to the y-axis is y²/a² - x²/b² = 1

- The coordinates of the vertices are  (0 , ± a)

- The coordinates of the co-vertices are (± b , 0)  

- The coordinates of the foci are (0 , ± c) where c² = a² + b²

* Lets solve the problem

∵ The equation of the hyperbola is y²/16 - x²/9 = 1

∵ The form of the equation is y²/a² - x²/b² = 1

∴ a² = 16

∴ a = √16 = 4

∵ The coordinates of the vertices are  (0 , ± a)

∴ The coordinates of the vertices are  (0 , 4) , (0 , -4)

∴ b² = 9

∴ b = √9 = 3

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = √25 = 5

∵ The coordinates of the foci are (0 , ± c)

∴ The coordinates of the foci are (0 , 5) , (0 , -5)

∵ The camera is pointed towards the vertex of the hyperbolic mirror

   which is (0 , 4) and is positioned such that the lens is at the nearest

  focus to that vertex which is (0 , 5)

∴ The distance between the lens and the mirror equal the distance

   between the vertex and the focus

∵ The vertex is (0 , 4) and the nearest focus is (0 , 5)

∴ The distance = 5 - 4 = 1 inch

* The lens is 1 inch from the mirror

the radius is 2 times the diameter.

step-by-step explanation: