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Write the equation for the hyperbola (below) in standard form and identify the vertices, foci and equations for the asymptotes. To earn full credit please share all steps and calculations. x^2+2x-100y^2-1000y+2401

Write the equation for the hyperbola (below) in standard form and identify the vertices-example-1
User McHat
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Answer:
((y+5)^2)/(7^2)-((x+1)^2)/(70^2)=1
\text{Foci = (}-1,\text{ -7}\sqrt[]{101}-5)\text{ and (}-1,\text{ -5+7}\sqrt[]{101})

The vertex = (-1, -12) and (-1, 2)


\begin{gathered} \text{First asymptote: }y=-(x+51)/(10) \\ \text{Second asymptote: }y=(x-49)/(10) \end{gathered}Step-by-step explanation:

The given equation of the hyperbola is:


x^2+2x-100y^2-1000y+2401

The standard form of the equation of a hyperbola is of the form


(\mleft(y-k\mright)^2)/(b^2)-((x-h)^2)/(a^2)=1

where (h, k) is the center of the hyperbola

The vertex = (h, k-b) and (h, k+b)

The given equation can be re-written as:


\begin{gathered} x^2+2x-100(y^2+10y)+2401=0 \\ x^2+2x-100(y^2+10y)=-2401 \\ (x^2+2x+1)-100(y^2+10y+25)=-2401+1-25(100) \\ (x^2+2x+1)-100(y^2+10y+25)=-4900 \\ 100(y^2+10y+25)-(x^2+2x+1)=4900 \end{gathered}

This can be further simplified into:


\begin{gathered} 100(y+5)^2-(x+1)^2=4900 \\ \text{Divide through by 100} \\ (y+5)^2-((x+1)^2)/(100)=(4900)/(100) \\ (y+5)^2-((x+1)^2)/(100)=49 \\ ((y+5)^2)/(49)-((x+1)^2)/(4900)=1 \\ ((y+5)^2)/(7^2)-((x+1)^2)/(70^2)=1 \end{gathered}

Therefore, the standard form of the given parabola equation is:


((y+5)^2)/(7^2)-((x+1)^2)/(70^2)=1

The center (h, k) = (-1, -5)

(a, b) = (70, 7)

The vertex = (-1, -5-7) and (-1, -5+7)

The vertex = (-1, -12) and (-1, 2)

The foci = (h, k-c) and (h, k+c)


\begin{gathered} c=\sqrt[]{a^2+b^2} \\ c=\sqrt[]{70^2+7^2} \\ c=7\sqrt[]{101} \end{gathered}
\text{Foci = (}-1,\text{ -7}\sqrt[]{101}-5)\text{ and (}-1,\text{ -5+7}\sqrt[]{101})

The first asymptote:


\begin{gathered} y=-(b)/(a)(x-h)+k \\ y=-(x+51)/(10) \\ \end{gathered}

Second asymptote


\begin{gathered} y=(b)/(a)(x-h)+k \\ y=(x-49)/(10) \end{gathered}

User Fdcpp
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