### Instruction: The hyperbola \[-x^2+2y^2-10x-16y+1=0\]has two foci. Find the coordinates of either of them. (Enter your answer as an ordered pair. Enter only one of the foci, not both.).### Response: To find the coordinates of the foci, I need to put the equation of the hyperbola in standard form, which looks like \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]or \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\] depending on whether the hyperbola is horizontal or vertical.
To do that, I need to complete the square for both x and y terms in the given equation.
I group the x and y terms together and factor out the coefficients of the squared terms: \[-x^2+2y^2-10x-16y+1=0 \implies -(x^2+10x)+2(y^2-8y)=-1\]
To complete the square for x, I need to add and subtract \(\left(\frac{10}{2}\right)^2=25\) inside the parentheses.
To complete the square for y, I need to add and subtract \(\left(\frac{8}{2}\right)^2=16\) inside the parentheses.
I also need to multiply the added and subtracted terms by the coefficients outside the parentheses, so that the equation does not change.
This gives me: \[-(x^2+10x+25-25)+2(y^2-8y+16-16)=-1 \implies -(x+5)^2+25+2(y-4)^2-32=-1\]
I simplify the constant terms and rearrange the equation: \[-(x+5)^2+2(y-4)^2-7=-1 \implies -(x+5)^2+2(y-4)^2=6\]
I divide both sides by 6 to get the standard form: \[\frac{-(x+5)^2}{6}+\frac{2(y-4)^2}{6}=1 \implies \frac{-(x+5)^2}{6}+\frac{(y-4)^2}{3}=1\]
Now I can identify the center, the semi-major axis, and the semi-minor axis of the hyperbola.
The center is \((h,k)=(-5,4)\).
The semi-major axis is \(a=\sqrt{3}\) since it corresponds to the positive term.
The semi-minor axis is \(b=\sqrt{6}\) since it corresponds to the negative term.
To find the foci, I need to use the relationship \(c^2=a^2+b^2\) for a hyperbola, where \(c\) is the distance from the center to either focus.
Plugging in the values of \(a\) and \(b\), I get \(c^2=3+6=9\) and \(c=3\).
Since the hyperbola is vertical, the foci are above and below the center by \(c\) units.
The coordinates of the foci are \((-5,4\pm 3)\) or \((-5,7)\) and \((-5,1)\).
I can choose either focus as my answer.