Answer:
The lens is 1 inch from the mirror
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