+0  
 
0
49
1
avatar

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
Sort: 

1+0 Answers

 #1
avatar+19207 
+1

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

18 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details