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Find the asymptote of -4x^2 + y^2 - 2y = 3 with positive slope. Enter your answer as an equation of the form y=mx+b.

Guest Mar 21, 2018
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Find the asymptote of -4x^2 + y^2 - 2y = 3 with positive slope. 

Enter your answer as an equation of the form y=mx+b.


Formula:
\(y^2 + Ax^2+Bx + Cxy +Dy + E = 0 \\ \text{$1.$ asymptote: $y = m_1x+b_1$} \\ \text{$2.$ asymptote: $y = m_2x+b_2$} \\ \begin{array}{|rcll|} \hline m_1 &=& \dfrac{-C+\sqrt{C^2-4A} }{2} \\ m_2 &=& \dfrac{-C-\sqrt{C^2-4A} }{2} \\ b_1 &=& \dfrac{B+m_1D}{m_2-m_1} \\ b_2 &=& -\dfrac{B+m_2D}{m_2-m_1} \\ \hline \end{array} \)

 

Equation of the hyperbola:  \(y^2 -4x^2 - 2y - 3 = 0\)

\(\begin{array}{|rcr|} \hline A &=& -4 \\ B &=& 0 \\ C &=& 0 \\ D &=& -2 \\ E &=& -3 \\ \hline \end{array} \begin{array}{|rcl|} \hline m_1 &=& \dfrac{-C+\sqrt{C^2-4A} }{2} \\ &=& \dfrac{0+\sqrt{0-4(-4)} }{2} \\ &=& \dfrac{ \sqrt{16} }{2} \\ &=& \dfrac{ 4 }{2} \\ \mathbf{m_1} & \mathbf{=} & \mathbf{2} \\ \hline m_2 &=& \dfrac{-C-\sqrt{C^2-4A} }{2} \\ &=& \dfrac{0-\sqrt{0-4(-4)} }{2} \\ &=& \dfrac{ -\sqrt{16} }{2} \\ &=& \dfrac{ -4 }{2} \\ \mathbf{m_2} & \mathbf{=} & \mathbf{-2} \\ \hline b_1 &=& \dfrac{B+m_1D}{m_2-m_1} \\ &=& \dfrac{0+2(-2)}{(-2)-2} \\ &=& \dfrac{-4}{-4} \\ \mathbf{b_1} & \mathbf{=} & \mathbf{1} \\ \hline b_2 &=& -\dfrac{B+m_2D}{m_2-m_1} \\ &=& -\dfrac{0+(-2)(-2)}{(-2)-2} \\ &=& -\dfrac{4}{-4} \\ \mathbf{b_2} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)


\(\text{$1.$ Asymptote: $y = 2x + 1$ (positive slope )}\\ \text{$2.$ Asymptote: $y = -2x +1$ } \)

 

laugh

heureka  Mar 23, 2018

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