We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
263
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.

 Mar 21, 2018
 #1
avatar+22152 
+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

 Mar 23, 2018

37 Online Users

avatar
avatar
avatar