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This printable supports Common Core Mathematics Standard HSG-GPE.A.3

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# Derivation of the Equation of a Hyperbola (Grades 11-12)

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## Derivation of the Equation of a Hyperbola

For the hyperbola centered at the origin with foci $F_1 = (-c,0)$ and $F_2=(c,0)$, derive the equation of the hyperbola.
1.
Which of the following is the best definition of a hyperbola?
1. Two opposing parabolas.
2. Given two points, called foci, the set of points such that the absolute value of the differences of the distance between the foci and a point is constant.
3. The distances between a set of points and the foci is constant.
4. Two points, foci, whose eccentricity is greater than 1.
2.
Let $(x,y)$ be a point on the hyperbola. What is the distance, $d_1$, between $F_1$ and the point $(x,y) ?$
1. $d_1 = sqrt( (x+c)^2 + y^2)$
2. $d_1 = sqrt( (x-c)^2 + y^2)$
3. $d_1 = sqrt( (x^2 + y^2)$
4. $d_1 = sqrt( (x-c)^2 + (y-c)^2)$
3.
What is the distance, $d_2$, between the $F_2$ and the point $(x,y) ?$
1. $d_2 = sqrt( (x-c)^2 + (y-c)^2)$
2. $d_2 = sqrt( x^2 + y^2)$
3. $d_2 = sqrt( (x+c)^2 + y^2)$
4. $d_2 = sqrt( (x-c)^2 + y^2)$
4.
Let $(a,0)$ be the right vertex of the hyperbola. What is the distance between this vertex and $F_1 ?$
1. $a-c$
2. $a+c$
3. $a$
4. $c$
5.
What is the distance between $F_2$ and the vertex $(a,0) ?$
1. $c$
2. $a$
3. $c-a$
4. $a+c$
6.
What does $|d_1 - d_2|$ equal?
1. $|d_1 - d_2| = a$
2. $|d_1 - d_2| = c$
3. $|d_1 - d_2| = |a-c|$
4. $|d_1 - d_2| = 2a$
7.
Substitute the equations for $d_1, d_2$ found in the second and third questions into the equation in the previous question. After some rearranging, the result is the following equation:

$sqrt( (x+c)^2 + y^2) = pm 2a + sqrt( (x-c)^2 + y^2)$

Why is there a plus-minus sign on the $2a$ term?
1. In order to remove the absolute value sign, one has to consider the case where $d_1>d_2$ and $d_1 < d_2$.
2. To remove the square root from this term, one has to look at the positive and negative case.
3. Since it's not certain which side of the equation this term should be on, it could be either positive or negative.
4. This is a hyperbola and the curves can be on either the positive or negative x-axis.
8.
Squaring both sides, expanding, and combing like terms, dividing by any common constants, and putting all terms without a square root on the left side of the equation results in which of the following equations?
1. $2cx - a = 0$
2. $cx pm a^2 = -2 sqrt((x-c)^2 + y^2)$
3. $cx - a^2 = 0$
4. $cx - a^2 = pm a sqrt( (x-c)^2 + y^2)$
9.
After some more algebra, the equation becomes:
$(c^2-a^2) x^2 - y^2 = (c^2 - a^2) y^2$
What happened to the plus-minus sign that was in the original equation in the seventh question?
1. Since there are no more square root terms, it is unnecessary.
2. Each term with it was eventually squared, and a squared term must be positive it.
3. Because the terms were moved from one side to another multiple times, it is no longer necessary.
4. Any terms with it canceled out.
10.
Letting $b^2 = c^2 - a^2$, the equation no becomes:
$b^2 x^2 - y^2 = b^2 a^2$
$x^2 / a^2 - y^2 / b^2 = 1$
This is the equation of a hyperbola, centered at the origin with foci $(-c,0)$ and $(c,0)$. Where does the substitution $b^2 = c^2 - a^2$ come from?
1. Since $b$ usually appears in the equation for a hyperbola, it must be included. Using the Pythagorean theorem, $a^2 +b^2 = c^2$, simply rearrange the equation.
2. It has to be done, to ensure the asymptotes are related to the equation. The equations of the asymptotes are $y= pm b/a$, and knowing that $|c| > |a|$, squaring and rearranging results in $b^2 = c^2 - a^2$.
3. It's done to simply the equation. $b$ is not defined yet, and since $|c| > |a|$, $c^2 > a^2$, and so there must be a positive number, $b^2$ such that $b^2 = c^2 - a^2$.
4. Knowing that $b$ is the length of the semi-minor axis, a right triangle can be formed with the center of the hyperbola and either foci, with $b$ as the length of one leg of this triangle. Applying the Pythagorean theorem results in $b^2 + c^2 = a^2$, and simply rearrange.        You need to be a HelpTeaching.com member to access free printables.