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Algebra II : Hyperbolic Functions

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Example Question #1 : Hyperbolic Functions

Which of the following equations represents a vertical hyperbola with a center of (-4, 3) and asymptotes at $y=\pm\, \frac{4}{3}\, x$ ?

Possible Answers:

$\frac{(y+3)^{2}}{16}-\frac{(x-4)^{2}}{9}=1$

$\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1$

$\frac{(x+4)^{2}}{9}-\frac{(y-3)^{2}}{16}=1$

$\frac{(y-3)^{2}}{9}-\frac{(x+4)^{2}}{16}=1$

Correct answer:

$\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1$

Explanation:

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

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

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

To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive $y^{2}$ term; a horizontal parabola has a positive $x^{2}$ term. In this case, we need a vertical parabola, so the $y^{2}$ 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 $y=\pm\, \frac{b}{a}\, x$ , where is found in the denominator of the $y^{2}$ term and is found in the denominator of the $x^{2}$ term. Since our asymptotes are $y=\pm\, \frac{4}{3}\, x$ , we know that must be 4 and must be 3. That means that the number underneath the $y^{2}$ term has to be 16, and the number underneath the $x^{2}$ term has to be 9.

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Example Question #1 : Hyperbolic Functions

What is the shape of the graph depicted by the equation:

$\small \frac{y^2}{9}-\frac{x^2}{4}=1$

Possible Answers:

Circle

Oval

Parabola

Hyperbola

Correct answer:

Hyperbola

Explanation:

The standard equation of a hyperbola is:

$\small \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

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Example Question #3 : Hyperbolic Functions

Express the following hyperbolic function in standard form:

13(y-7)^2-13(x+6)^2=52

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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:

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

Simplifying, we obtain our final answer in standard form:

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

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Example Question #1 : Hyperbolic Functions

Which of the following answers best represent $2\sinh(x)$ ?

Possible Answers:

$\frac{e^x-e^-^x}{2}$

$\frac{2}{e^x+e^-^x}$

e^x+e^-^x

e^x-e^-^x

$\frac{e^x+e^-^x}{2}$

Correct answer:

e^x-e^-^x

Explanation:

The correct definition of hyperbolic sine is:

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

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

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

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Example Question #5 : Hyperbolic Functions

Which of the following best represents $\cosh(x)-\sinh(x)$ , if the value of is zero?

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

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

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

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

Therefore:

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

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Example Question #6 : Hyperbolic Functions

What is the value of $-\tanh(0)$ ?

Possible Answers:

$\frac{1}{2}$

$\text{Does not exist}$

Correct answer:

Explanation:

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

According to the properties:

$\sinh(0)=0$

$\cosh(0)=1$

Therefore:

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

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Example Question #7 : Hyperbolic Functions

Simplify: $-\frac{1}{\sinh^2(x)-\cosh^2(x)}$

Possible Answers:

x^2

$\text{Does not exist}$

Correct answer:

Explanation:

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

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

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

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

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

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

Substitute the value into the problem.

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

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Example Question #8 : Hyperbolic Functions

$\small f(x)=\tfrac{1}{x+4}-3$

Which of the following is the correct expression for a hyperbola that is shifted $\small 5$ $\small units$ units up and $\small 2$ $\small units$ to the right of $\small f(x)$ ?

Possible Answers:

$\small \small \small \small g(x)=\tfrac{1}{x+2}+5$

$\small \small \small g(x)=\tfrac{1}{x+2}+2$

$\small \small \small \small g(x)=\tfrac{1}{x+9}-1$

$\small \small \small \small g(x)=\tfrac{1}{x-2}+5$

$\small \small \small \small g(x)=\tfrac{1}{x+6}-8$

Correct answer:

$\small \small \small g(x)=\tfrac{1}{x+2}+2$

Explanation:

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

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Example Question #9 : Hyperbolic Functions

Find the foci of the hyperbola: $-\frac{x^2}{36}+\frac{y^2}{16} =1$

Possible Answers:

$(\pm 2\sqrt{13},0)$

$(0,\pm 2\sqrt{13})$

$(\pm6,0)$

$(\pm 4\sqrt{13},0)$

$(0,\pm4)$

Correct answer:

$(0,\pm 2\sqrt{13})$

Explanation:

Write the standard forms for a hyperbola.

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

OR:

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

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

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

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

$-\frac{x^2}{36}+\frac{y^2}{16} =1$

Identify the coefficients a,b and substitute to find the value of .

$a^2 = 36, a=\pm6$

$b^2 =16, b=\pm 4$

$c^2 =a^2+b^2\rightarrow c^2 =36+16$

$c^2=52, c=\pm \sqrt{52} =\pm \sqrt{4}\cdot \sqrt{13}= \pm2\sqrt{13}$

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

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

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Example Question #10 : Hyperbolic Functions

Given the hyperbola y^2 +4y = 4x^2-24x+41 , what is the value of the center?

Possible Answers:

Correct answer:

Explanation:

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

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

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

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

The equation becomes:

y^2+4y-4x^2+24x = 41

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

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

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

(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.

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

This equation becomes:

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

Factorize the left side and simplify the right.

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

Divide both sides by nine.

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

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

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

The center is at: (3,-2)

The answer is:

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Algebra II : Hyperbolic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

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