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.

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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.

The answer is:[tex] \frac{ y^{2} }{256} } - \frac{ x^{2} }{ 144 } = 1[/tex]Explanation:The general equation for a hyperbola with transverse axis vertical is[tex] \frac{ (y - k)^{2} }{ a^{2} } - \frac{ (x - h)^{2} }{ b^{2} } = 1[/tex]where:h = x-coordinate of the centerk = y-coordinate of the centera = semi-major axisb = semi-minor axisWe know that the center has coordinates (0, 0), therefore:h = 0k = 0We also know thata = 162c = distance between foci = 40In order to find b², we need to use the formula c² = a² + b²and solve it for b²:b² = c² - a² = (40÷2)² - 16² = 400 - 256 = 144Now we can substitute the obtained values in the general equation to find the required equation:[tex] \frac{ y^{2} }{256} } - \frac{ x^{2} }{ 144 } = 1[/tex]

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