First, we need to become familiar with the standard form of a hyperbolic equation:

\(\displaystyle \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)

The center is always at \(\displaystyle (h,k)\). This means that for this problem, the numerators of each term will have to contain \(\displaystyle (x+4)^{2}\) and \(\displaystyle (y-3)^{2}\).

To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive \(\displaystyle y^{2}\) term; a horizontal parabola has a positive \(\displaystyle x^{2}\) term. In this case, we need a vertical parabola, so the \(\displaystyle y^{2}\) term will have to be positive.

(NOTE: If both terms are the same sign, you have an ellipse, not a parabola.)

The asymptotes of a parabola are always found by the equation \(\displaystyle y=\pm\, \frac{b}{a}\, x\), where \(\displaystyle b\) is found in the denominator of the \(\displaystyle y^{2}\) term and \(\displaystyle a\) is found in the denominator of the \(\displaystyle x^{2}\) term. Since our asymptotes are \(\displaystyle y=\pm\, \frac{4}{3}\, x\), we know that \(\displaystyle b\) must be 4 and \(\displaystyle a\) must be 3. That means that the number underneath the \(\displaystyle y^{2}\) term has to be 16, and the number underneath the \(\displaystyle x^{2}\) term has to be 9.

The standard equation of a hyperbola is:

\(\displaystyle \small \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\)

In order to express the given hyperbolic function in standard form, we must write it in one of the following two ways:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

\(\displaystyle \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\)

From our formulas for the standard form of a hyperbolic equation above, we can see that the term on the right side of the equation is always 1, so we must divide both sides of the given equation by 52, which gives us:

\(\displaystyle \frac{13(y-7)^2}{52}-\frac{13(x+6)^2}{52}=1\)

Simplifying, we obtain our final answer in standard form:

\(\displaystyle \frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1\)

The correct definition of hyperbolic sine is:

\(\displaystyle \sinh(x)= \frac{e^x-e^-^x}{2}\)

Therefore, by multiplying 2 by both sides we get the following answer,

\(\displaystyle 2\sinh(x)=e^x-e^-^x\)

Find the values of hyperbolic sine and cosine when x is zero.  According to the properties:

\(\displaystyle \sinh(0)=\frac{e^x-e^{-x}}{2}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0\)

\(\displaystyle \cosh(0)= \frac{e^x+e^{-x}}{2}=\frac{e^0+e^{-0}}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)

Therefore:

\(\displaystyle \cosh(0)-\sinh(0)=1-0=1\)

The hyperbolic tangent will need to be rewritten in terms of hyperbolic sine and cosine.

According to the properties:

\(\displaystyle \sinh(0)=0\)

\(\displaystyle \cosh(0)=1\)

Therefore:

\(\displaystyle -\tanh(0)= -\frac{\sinh(0)}{\cosh(0)}=-\frac{\frac{e^0-e^0}{2}}{\frac{e^0+e^0}{2}}=-\frac{e^0-e^0}{e^0+e^0}=-\frac{0}{1} = 0\)

The following is a property of hyperbolics that is closely similar to the problem.

\(\displaystyle \cosh^2(x) -\sinh^2(x)= 1\)

We will need to rewrite this equation by taking a negative one as the common factor, and divide the negative one on both sides. 

\(\displaystyle (-1)(-\cosh^2(x) +\sinh^2(x))= 1\)

\(\displaystyle -\cosh^2(x) +\sinh^2(x)= -1\)

\(\displaystyle \sinh^2(x)-\cosh^2(x) =-1\)

Substitute the value into the problem.

\(\displaystyle -\frac{1}{sinh^2(x)-cosh^2(x)} = - \frac{1}{-1} =1\)

The parent function of a hyperbola is represented by the function \(\displaystyle \small \small \small \small g(x)=\tfrac{1}{x-h}+k\) where \(\displaystyle \small (h,k)\) is the center of the hyperbola. To shift the original function up by \(\displaystyle \small 5\) \(\displaystyle \small units\) simply add \(\displaystyle \small -3+5\). To shift it to the right \(\displaystyle \small 2\) \(\displaystyle \small units\) take away \(\displaystyle \small 4-2\).

Write the standard forms for a hyperbola.

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} =1\)

OR:

\(\displaystyle -\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} =1\)

The standard form is given in the second case, which will have different parameters compared to the first form.

Center:  \(\displaystyle (h,k) =(0,0)\)

Foci:  \(\displaystyle (h,k\pm c)\), where \(\displaystyle c^2 =a^2+b^2\)

\(\displaystyle -\frac{x^2}{36}+\frac{y^2}{16} =1\)

Identify the coefficients \(\displaystyle a,b\) and substitute to find the value of \(\displaystyle c\).

\(\displaystyle a^2 = 36, a=\pm6\)

\(\displaystyle b^2 =16, b=\pm 4\)

\(\displaystyle c^2 =a^2+b^2\rightarrow c^2 =36+16\)

\(\displaystyle c^2=52, c=\pm \sqrt{52} =\pm \sqrt{4}\cdot \sqrt{13}= \pm2\sqrt{13}\)

\(\displaystyle (h,k\pm c) = (0,0 \pm 2\sqrt{13}) = (0,\pm 2\sqrt{13})\)

The answer is:  \(\displaystyle (0,\pm 2\sqrt{13})\)

In order to determine the center, we will first need to rewrite this equation in standard form.  

\(\displaystyle \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\)

Isolate 41 on the right side.  Subtract \(\displaystyle 4x^2\) and add \(\displaystyle 24x\) on both sides.

\(\displaystyle y^2 +4y -(4x^2)+[24x]= 4x^2-24x+41-(4x^2)+[24x]\)

The equation becomes:

\(\displaystyle y^2+4y-4x^2+24x = 41\)

Group the x and y terms.  Be careful of the negative signs.

\(\displaystyle (y^2+4y)-(4x^2-24x )= 41\)

Pull out a common factor of 4 on the second parentheses.

\(\displaystyle (y^2+4y)-4(x^2-6x )= 41\)

Complete the square twice.  Divide the second term of each parentheses by two and square the quantity.  Add the terms on both sides.

\(\displaystyle (y^2+4y+(\frac{4}{2})^2)-4(x^2-6x+(\frac{-6}{3})^2)= 41+(\frac{4}{2})^2+(-4)(\frac{-6}{3})^2\)

This equation becomes:

\(\displaystyle (y^2+4y+4)-4(x^2-6x+9) = 41+4-36\)

Factorize the left side and simplify the right.

\(\displaystyle (y+2^2)-4(x-3)^2 = 9\)

Divide both sides by nine.

\(\displaystyle \frac{(y+2)^2}{9}-\frac{4}{9}(x-3)^2 = \frac{9}{9}\)

\(\displaystyle \frac{(y+2)^2}{9}-\frac{4(x-3)^2 }{9}=1\)

The equation is now in the standard form of a hyperbola.

The center is at:  \(\displaystyle (3,-2)\)

The answer is:  \(\displaystyle 3\)