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The asymptotes of a hyperbola are \(y = 2x - 3\) and \(y = 17 - 2x. \) Also, the hyperbola passes through the point \((4,7).\) Find the distance between the foci of the hyperbola.

 Aug 5, 2019
 #1
avatar+23293 
+3

The asymptotes of a hyperbola are \(y = 2x - 3\) and \(y = 17 - 2x\).
Also, the hyperbola passes through the point \((4,7)\).
Find the distance between the foci of the hyperbola.

 

Formula hyperbola:

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ asymptote}} & 2x - 3 &=& y \\ & 2x - 3-y &=& 0 \\\\ \hline \mathbf{2. \text{ asymptote}} & 17 - 2x &=& y \\ & 17 - 2x-y &=& 0 \\\\ \hline \text{hyperbola: } & (2x - 3-y)(17 - 2x-y) + c &=& 0 \\\\ P(x=4,y=7) & (8-3-7)(17-8-7)+c &=& 0 \\ & (-2)(2)+c &=& 0 \\ & c &=& 4 \\ \mathbf{\text{hyperbola: }} &\mathbf{ (2x - 3-y)(17 - 2x-y) + 4} &\mathbf{=}& \mathbf{0} \\\\ & (2x - 3-y)(17 - 2x-y) + 4 &=& 0 \\ & 34x-4x^2-2xy-51+6x+3y-17y+2xy+y^2+4 &=& 0 \\ & -4x^2+40x+y^2-14y &=& 47 \\ &\ldots \\ & \mathbf{ \dfrac{(x-5)^2}{1} - \dfrac{(y-7)^2}{4}} &=& \mathbf{ 1 } \\ & \boxed{\mathbf{\text{Formula hyperbola:} } \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\\\ & a^2 = 1 \\ & \mathbf{a= 1} \\\\ & b^2 = 4 \\ & \mathbf{b= 2} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=&\mathbf{ 2 \sqrt{a^2+b^2}} \quad | \quad a^2=1, \quad b^2= 4 \\ & &=& 2 \sqrt{1+4} \\ & &\mathbf{=}& \mathbf{2\sqrt{5} } \\ \hline \end{array} \)

 

The distance between the foci of the hyperbola is \(\mathbf{2\sqrt{5} }\)

 

laugh

 Aug 5, 2019
 #2
avatar+928 
+1

Wow heureka! You have a greatly presented solution!!!

dgfgrafgdfge111  Aug 5, 2019
 #3
avatar+23293 
+1

Thank you dgfgrafgdfge111

 

laugh

heureka  Aug 5, 2019

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