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Find the distance between the vertices of the hyperbola \(9x^2 + 54x - y^2 + 10y + 55 = 0.\)

 Apr 9, 2019

Best Answer 

 #1
avatar+6196 
+2

\(9x^2 +54x - (y^2 - 10y)+55 = \\ 9(x^2 + 6x +9 - 9)-(y^2-10y +25-25) + 55=\\ 9(x+3)^2 - 81 - (y-5)^2 +25 + 55 = \\ 9(x+3)^2 - (y-5)^2 = 1\\ \dfrac{(x+3)^2}{\left(\frac 1 3\right)^2} - (y-5)^2 = 1\)

 

\(\text{The vertices are at }\\ (x,y)=(-3\pm \dfrac 1 3 ,5)=\left (-\dfrac{10}{3},5\right),~\left(-\dfrac{8}{3},5\right)\\ \text{and thus they are }\dfrac 2 3 \text{ units apart}\)

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 Apr 9, 2019
 #1
avatar+6196 
+2
Best Answer

\(9x^2 +54x - (y^2 - 10y)+55 = \\ 9(x^2 + 6x +9 - 9)-(y^2-10y +25-25) + 55=\\ 9(x+3)^2 - 81 - (y-5)^2 +25 + 55 = \\ 9(x+3)^2 - (y-5)^2 = 1\\ \dfrac{(x+3)^2}{\left(\frac 1 3\right)^2} - (y-5)^2 = 1\)

 

\(\text{The vertices are at }\\ (x,y)=(-3\pm \dfrac 1 3 ,5)=\left (-\dfrac{10}{3},5\right),~\left(-\dfrac{8}{3},5\right)\\ \text{and thus they are }\dfrac 2 3 \text{ units apart}\)

Rom Apr 9, 2019

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