Find the location of the center, vertices, and foci for the hyperbola described by the equation.
$$\frac{ (x+4)^2 }{ 36 } - \frac{ (y-1)^2 }{ 25 }$$

Question

Find the location of the center, vertices, and foci for the hyperbola described by the equation. 
$$\frac{ (x+4)^2 }{ 36 } - \frac{ (y-1)^2 }{ 25 }$$

anonymous · Answer

$$\frac{ (x+4)^2 }{ 36 } - \frac{ (y-1)^2 }{ 25} = 1$$ **

anonymous · Answer

center (-4,1)

imstuck · Answer

The vertices are a = 6 and b = 5, and since this is an x^2 parabola, the foci and the vertices will lie on the x axis.  So the vertices are: (2,1) and (-10, 1)

imstuck · Answer

The formula for the foci is a^2 + b^2 = c^2. So in translation that is:$$36+25=c ^{2}$$$$61=c ^{2}$$and $$c=\sqrt{61}$$But of course we have to move those according to the center. The foci also lie on the x axis, so the coordinates of the foci are$$(-4+\sqrt{61},1)$$and $$(-4-\sqrt{61},1)$$

imstuck · Answer

So this is what your plane would look like with those points plotted:

anonymous · Answer

Thank you very much and I have a question...where did the 4 come from?

anonymous · Answer

-4*

imstuck · Answer

hold on.  let me check something, ok?

anonymous · Answer

okay

anonymous · Answer

|dw:1406043003342:dw|

imstuck · Answer

That's an ellipse, @ramit.dour

anonymous · Answer

Oh !!! sorry.

anonymous · Answer

lol it's okay thanks for trying @ramit.dour

anonymous · Answer

|dw:1406043148731:dw|

imstuck · Answer

sorry I'm trying to do this...distractions here at home...blah blah blah...let me work on this for a sec.

anonymous · Answer

@IMStuck take your time, I appreciate it.
@ramit.dour is the (-4,1) supposed to be the center point of the two curves? o.o

imstuck · Answer

ok, here it is:

imstuck · Answer

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