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3 years ago

Help me with this please

Mathematics

2 answers:

Vesnalui [34]3 years ago

6 0

I think it's -6r/s but I'm not sure

Maurinko [17]3 years ago

3 0

The answer is $\frac{-6r}{s}$

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3 years ago

Read 2 more answers

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

Ksivusya [100]

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)

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2 years ago