Question

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 1

Answer:

$\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1$.

Step-by-step explanation:

If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is $\frac{(x - \alpha )^{2} }{a^{2} } - \frac{(y - \beta )^{2} }{b^{2} } = 1$.

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 $b^{2} = a^{2}(e^{2} - 1) = 4[(\frac{5}{2} )^{2} -1] = 21$ {Since $e = \frac{5}{2}$ given}

Therefore, the equation of the given hyperbola will be  

$\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1$. (Answer)