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A hyperbola centered at the origin has foci at (\(\pm 7\), 0), and passes through the point (2, 12). If the equation of the hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\)where a and b are positive constants, compute the ordered pair (a, b).

 Jul 28, 2019

Best Answer 

 #1
avatar+23082 
+3

A hyperbola centered at the origin has foci at \((\pm 7, 0)\), and passes through the point \((2, 12)\).
If the equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), where \(a\) and \(b\) are positive constants,
compute the ordered pair \((a, b)\).

 

\(\begin{array}{|lrcll|} \hline P(2,12): & \mathbf{\dfrac{2^2}{a^2} - \dfrac{12^2}{b^2}} &=& \mathbf{1} \\\\ & \dfrac{4}{a^2} - \dfrac{144}{b^2}&=& 1 \\\\ & 4b^2- 144a^2 &=& a^2b^2 \\ & 144a^2 &=& 4b^2-a^2b^2 \\ & 144a^2 &=& b^2(4-a^2) \quad | \quad a^2+b^2 = 7^2,\ \mathbf{b^2=49-a^2} \\ & 144a^2 &=& (49-a^2)(4-a^2) \\ & 144a^2 &=& 196-53a^2+a^4 \\ & 197a^2 &=& 196+a^4 \\ & a^4 - 197a^2+196 &=& 0 \\\\ & a^2 &=& \dfrac{197\pm\sqrt{197^2-4\cdot 196}} {2} \\ & a^2 &=& \dfrac{197\pm\sqrt{38025}} {2} \\ & a^2 &=& \dfrac{197\pm 195} {2} \\\\ & a^2 &=& \dfrac{197 + 195} {2} \\ & a^2 &=& 196, \ b^2 = 49-196 \quad \text{no solution $b$ is complex } \\\\ & a^2 &=& \dfrac{197 - 195} {2} \\ & a^2 &=& 1, \ b^2 = 49-1=48 \ \checkmark \\ & \mathbf{a} &=& \mathbf{1}, \ \mathbf{ b}= \sqrt{48} \mathbf{= 4\sqrt{3}} \\ \hline \end{array}\)

 

\(\mathbf{ (a,b) = (1,\ 4\sqrt{3}) }\)

 

laugh

 Jul 29, 2019
 #1
avatar+23082 
+3
Best Answer

A hyperbola centered at the origin has foci at \((\pm 7, 0)\), and passes through the point \((2, 12)\).
If the equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), where \(a\) and \(b\) are positive constants,
compute the ordered pair \((a, b)\).

 

\(\begin{array}{|lrcll|} \hline P(2,12): & \mathbf{\dfrac{2^2}{a^2} - \dfrac{12^2}{b^2}} &=& \mathbf{1} \\\\ & \dfrac{4}{a^2} - \dfrac{144}{b^2}&=& 1 \\\\ & 4b^2- 144a^2 &=& a^2b^2 \\ & 144a^2 &=& 4b^2-a^2b^2 \\ & 144a^2 &=& b^2(4-a^2) \quad | \quad a^2+b^2 = 7^2,\ \mathbf{b^2=49-a^2} \\ & 144a^2 &=& (49-a^2)(4-a^2) \\ & 144a^2 &=& 196-53a^2+a^4 \\ & 197a^2 &=& 196+a^4 \\ & a^4 - 197a^2+196 &=& 0 \\\\ & a^2 &=& \dfrac{197\pm\sqrt{197^2-4\cdot 196}} {2} \\ & a^2 &=& \dfrac{197\pm\sqrt{38025}} {2} \\ & a^2 &=& \dfrac{197\pm 195} {2} \\\\ & a^2 &=& \dfrac{197 + 195} {2} \\ & a^2 &=& 196, \ b^2 = 49-196 \quad \text{no solution $b$ is complex } \\\\ & a^2 &=& \dfrac{197 - 195} {2} \\ & a^2 &=& 1, \ b^2 = 49-1=48 \ \checkmark \\ & \mathbf{a} &=& \mathbf{1}, \ \mathbf{ b}= \sqrt{48} \mathbf{= 4\sqrt{3}} \\ \hline \end{array}\)

 

\(\mathbf{ (a,b) = (1,\ 4\sqrt{3}) }\)

 

laugh

heureka Jul 29, 2019

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