the eccentricity of a hyperbola is defined as e=c/a. find an equation with vertices (1,-3) and (-3,-3) and e=5/2

Answer:[tex]\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1[/tex].Step-by-step explanation:If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is [tex]\frac{(x - \alpha )^{2} }{a^{2} } - \frac{(y - \beta )^{2} }{b^{2} } = 1[/tex]. Now, the vertices of the hyperbola are given by (α ± a, β) ≡ (1,-3) and (-3,-3) Hence, β = - 3 and α + a = 1 and α - a = -3 Now, solving those two equations of α and a we get,  2α = - 2, ⇒ α = -1 and a = 1 - α = 2. Now, eccentricity of the hyperbola is given by [tex]b^{2} = a^{2}(e^{2}  - 1) = 4[(\frac{5}{2} )^{2} -1] = 21[/tex] {Since [tex]e = \frac{5}{2}[/tex] given} Therefore, the equation of the given hyperbola will be  [tex]\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1[/tex]. (Answer)