Question

Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1

Answer 1

Answer:

I don't have the full answer but one of them for sure is:

$\frac{(x-12)^2}{4^2} - \frac{(y+28)^2}{3^2} =1$

Step-by-step explanation:

Answer 2

$\bf \textit{hyperbolas, vertical traverse axis }\\\\ \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}, {{ k}}\pm a) \end{cases}\\\\ -------------------------------\\\\ \cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\ -------------------------------\\\\ c=\textit{distance from the center to either foci}\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}$

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)