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The ellipse \(x^2 +4y^2 =4\) and the hyperbola \(x^2 -m(y+2)^2 =1\) are tangent. Compute \(m\)

 

Thanks in advance.

 Aug 10, 2020
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The ellipse \(x^2 +4y^2 =4\) and the hyperbola \(x^2 -m(y+2)^2 =1\) are tangent.

Compute \(m\)

 

My attempt

 

1. The coordinate of the ellipse and the hyperbola must be the same

\(\begin{array}{|rcll|} \hline \mathbf{x^2+4y^2} &=& \mathbf{4} \\ \mathbf{x^2} &=& \mathbf{4-4y^2} \\ \hline \mathbf{x^2 -m(y+2)^2} &=& \mathbf{1} \quad | \quad \mathbf{x^2=4-4y^2} \\ \mathbf{4-4y^2 -m(y+2)^2} &=& \mathbf{1} \qquad (1) \\ \hline \end{array}\)

 

2. The slope at the ellipse and at the hyperbola must be equal

\(\begin{array}{|rcll|} \hline \text{The slope at the ellipse} \\ \hline \dfrac{\partial f(x,y)} {\partial x} &=& 2x \\ \dfrac{\partial f(x,y)} {\partial y} &=& 8y \\ && \boxed{\text{Formula: }\\ \dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f} {\partial x}} {\dfrac{\partial f} {\partial y}} } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ 2x } { 8y } \\\\ \mathbf{ \dfrac{dy}{dx}} &\mathbf{=}& \mathbf{-\dfrac{ x } { 4y }} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{The slope at the hyperbola} \\ \hline \dfrac{\partial f(x,y)} {\partial x} &=& 2x \\ \dfrac{\partial f(x,y)} {\partial y} &=& -2m(y+2)*1 \\ && \boxed{\text{Formula: }\\ \dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f} {\partial x}} {\dfrac{\partial f} {\partial y}} } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ 2x } { -2m(y+2) } \\\\ \mathbf{ \dfrac{dy}{dx}} &\mathbf{=}& \mathbf{\dfrac{ x } {m(y+2) }} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{ \dfrac{dy}{dx}} = \mathbf{-\dfrac{ x } { 4y }} &\mathbf{=}& \mathbf{\dfrac{ x } {m(y+2) }}\\ -\dfrac{ x } { 4y } &=& \dfrac{ x } {m(y+2) }\\ -\dfrac{ 1 } { 4y } &=& \dfrac{ 1 } {m(y+2) }\\ m(y+2) &=& -4y \\ \mathbf{m} &=& \mathbf{-\dfrac{4y}{y+2} } \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{4-4y^2 -m(y+2)^2} &=& \mathbf{1} \quad | \quad \mathbf{m=-\dfrac{4y}{y+2} } \\ & 4-4y^2 +\dfrac{4y}{(y+2)}(y+2)^2 &=& 1 \\ & 4-4y^2 +4y(y+2) &=& 1 \\ & 4-4y^2 +4y^2+8y &=& 1 \\ & 4 +8y &=& 1 \\ & 8y &=& -3 \\ & \mathbf{y} &=&\mathbf{ -\dfrac{3}{8}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{m} &=& \mathbf{-\dfrac{4y}{y+2} } \quad | \quad \mathbf{y=-\dfrac{3}{8}} \\\\ m &=& -\dfrac{4\left(-\dfrac{3}{8}\right) }{-\dfrac{3}{8}+2} \\\\ m &=& \dfrac{ \dfrac{3}{2} }{2-\dfrac{3}{8}} \\\\ m &=& \dfrac{3}{2}\above 1pt \dfrac{13}{8} \\ \\ m &=& \dfrac{3}{2} * \dfrac{8}{13} \\\\ \mathbf{m} &=& \mathbf{\dfrac{12}{13}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x^2} &=& \mathbf{4-4y^2} \quad | \quad \mathbf{y=-\dfrac{3}{8}} \\\\ x^2&=& 4-4\left(\dfrac{3}{8}\right)^2 \\\\ x^2&=& 4-\dfrac{4*9}{64} \\\\ x^2&=& 4-\dfrac{9}{16} \\\\ x^2&=& \dfrac{64-9}{16} \\\\ x^2&=& \dfrac{55}{16} \\\\ \mathbf{x} &=& \mathbf{ \dfrac{\sqrt{55}}{4} } \\ \hline \end{array}\)

 

 

wink

 Aug 11, 2020
edited by heureka  Aug 11, 2020

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