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I don't get eccentricity and its relation to hyperbolas...in a question where I have to form an equation for a hyperbola with a center at (-3,1), one focus at (2,1), and eccentricity is 5/4, how would I solve this? 

Guest Jan 22, 2018
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I don't get eccentricity and its relation to hyperbolas...in a question where I have to form an equation for a hyperbola with a center at (-3,1), one focus at (2,1), and eccentricity is 5/4, how would I solve this? 

 

\(center : ~(h=-3,~ k=1)\)

\(focus: ~( x_f = 2,~ y_f=1)\)

\(eccentricity: ~e = \frac{5}{4}\)

 

Because \(k = y_f =1\) the transverse axis is horizontal.

 

1.  c = ?

\(\begin{array}{|rcll|} \hline focus: ( h+c,~k ) &=& (2,1)\\ h+c &=& 2 \\ -3+c &=& 2 \\ c&=& 2+3\\ \mathbf{c} &\mathbf{=}& \mathbf{5} \\ \hline \end{array}\)

 

(2.)

\(\begin{array}{|rcll|} \hline focus_2: ( h-c,~k ) &=& (x_f,y_f=k)\\ h-c &=& x_f \\ -3-5 &=& x_f \\ x_f &=& -8\\ y_f &=& 1 \\ \mathbf{focus_2} &\mathbf{=}& \mathbf{(-8,1)} \\ \hline \end{array} \)

 

3. a = ?

\(\begin{array}{|rcll|} \hline \mathbf{e} &\mathbf{=}& \mathbf{\frac{c}{a}} \\ \frac{5}{4} &=& \frac{5}{a} \\ \frac{4}{5} &=& \frac{a}{5} \\ a &=& \frac{4}{5}\cdot 5 \\ \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

 

4. b = ?

\(\begin{array}{|rcll|} \hline \mathbf{c^2} &\mathbf{=}& \mathbf{a^2+b^2} \\ b^2 &=&c^2-a^2 \\ b^2 &=&5^2-4^2 \\ b^2 &=&25-16 \\ b^2 &=& 9 \\ \mathbf{b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)

 

5.  Equation:

\(\begin{array}{|rcll|} \hline \mathbf{\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}} &\mathbf{=}& \mathbf{1} \\ \frac{(x-(-3))^2}{4^2}-\frac{(y-1)^2}{3^2} &=& 1 \\ \mathbf{\frac{(x+3)^2}{4^2}-\frac{(y-1)^2}{3^2}} &\mathbf{=}& \mathbf{1} \\ \hline \end{array} \)

 

6. The graph:

 

 

laugh

heureka  Jan 22, 2018

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