The vertex = (-1, -12) and (-1, 2)
[tex]\begin{gathered} \text{First asymptote: }y=-\frac{x+51}{10} \\ \text{Second asymptote: }y=\frac{x-49}{10} \end{gathered}[/tex]Explanation:
The given equation of the hyperbola is:
[tex]x^2+2x-100y^2-1000y+2401[/tex]The standard form of the equation of a hyperbola is of the form
[tex]\frac{\mleft(y-k\mright)^2}{b^2}-\frac{(x-h)^2}{a^2}=1[/tex]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:
[tex]\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}[/tex]This can be further simplified into:
[tex]\begin{gathered} 100(y+5)^2-(x+1)^2=4900 \\ \text{Divide through by 100} \\ (y+5)^2-\frac{(x+1)^2}{100}=\frac{4900}{100} \\ (y+5)^2-\frac{(x+1)^2}{100}=49 \\ \frac{(y+5)^2}{49}-\frac{(x+1)^2}{4900}=1 \\ \frac{(y+5)^2}{7^2}-\frac{(x+1)^2}{70^2}=1 \end{gathered}[/tex]Therefore, the standard form of the given parabola equation is:
[tex]\frac{(y+5)^2}{7^2}-\frac{(x+1)^2}{70^2}=1[/tex]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)
[tex]\begin{gathered} c=\sqrt[]{a^2+b^2} \\ c=\sqrt[]{70^2+7^2} \\ c=7\sqrt[]{101} \end{gathered}[/tex][tex]\text{Foci = (}-1,\text{ -7}\sqrt[]{101}-5)\text{ and (}-1,\text{ -5+7}\sqrt[]{101})[/tex]The first asymptote:
[tex]\begin{gathered} y=-\frac{b}{a}(x-h)+k \\ y=-\frac{x+51}{10} \\ \end{gathered}[/tex]Second asymptote
[tex]\begin{gathered} y=\frac{b}{a}(x-h)+k \\ y=\frac{x-49}{10} \end{gathered}[/tex]