# 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 [math]F_1 = (-c,0)[/math] and [math]F_2=(c,0)[/math], derive the equation of the hyperbola.

1.

Which of the following is the best definition of a hyperbola?

- Two opposing parabolas.
- 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.
- The distances between a set of points and the foci is constant.
- Two points, foci, whose eccentricity is greater than 1.

2.

Let [math](x,y)[/math] be a point on the hyperbola. What is the distance, [math]d_1[/math], between [math]F_1[/math] and the point [math](x,y) ?[/math]

- [math] d_1 = sqrt( (x+c)^2 + y^2) [/math]
- [math] d_1 = sqrt( (x-c)^2 + y^2)[/math]
- [math] d_1 = sqrt( (x^2 + y^2) [/math]
- [math]d_1 = sqrt( (x-c)^2 + (y-c)^2)[/math]

3.

What is the distance, [math]d_2[/math], between the [math]F_2[/math] and the point [math](x,y) ?[/math]

- [math]d_2 = sqrt( (x-c)^2 + (y-c)^2) [/math]
- [math]d_2 = sqrt( x^2 + y^2)[/math]
- [math]d_2 = sqrt( (x+c)^2 + y^2)[/math]
- [math]d_2 = sqrt( (x-c)^2 + y^2)[/math]

4.

Let [math](a,0)[/math] be the right vertex of the hyperbola. What is the distance between this vertex and [math]F_1 ?[/math]

- [math]a-c[/math]
- [math]a+c[/math]
- [math]a[/math]
- [math]c[/math]

5.

What is the distance between [math]F_2[/math] and the vertex [math](a,0) ?[/math]

- [math]c[/math]
- [math]a[/math]
- [math]c-a[/math]
- [math]a+c[/math]

6.

What does [math]|d_1 - d_2|[/math] equal?

- [math]|d_1 - d_2| = a[/math]
- [math]|d_1 - d_2| = c[/math]
- [math]|d_1 - d_2| = |a-c| [/math]
- [math]|d_1 - d_2| = 2a[/math]

7.

Substitute the equations for [math]d_1, d_2[/math] found in the second and third questions into the equation in the previous question. After some rearranging, the result is the following equation:

[math]sqrt( (x+c)^2 + y^2) = pm 2a + sqrt( (x-c)^2 + y^2)[/math]

Why is there a plus-minus sign on the [math]2a[/math] term?

[math]sqrt( (x+c)^2 + y^2) = pm 2a + sqrt( (x-c)^2 + y^2)[/math]

Why is there a plus-minus sign on the [math]2a[/math] term?

- In order to remove the absolute value sign, one has to consider the case where [math]d_1>d_2[/math] and [math]d_1 < d_2[/math].
- To remove the square root from this term, one has to look at the positive and negative case.
- Since it's not certain which side of the equation this term should be on, it could be either positive or negative.
- 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?

- [math]2cx - a = 0[/math]
- [math]cx pm a^2 = -2 sqrt((x-c)^2 + y^2) [/math]
- [math]cx - a^2 = 0[/math]
- [math]cx - a^2 = pm a sqrt( (x-c)^2 + y^2) [/math]

9.

After some more algebra, the equation becomes:

[math](c^2-a^2) x^2 - y^2 = (c^2 - a^2) y^2[/math]

What happened to the plus-minus sign that was in the original equation in the seventh question?

[math](c^2-a^2) x^2 - y^2 = (c^2 - a^2) y^2[/math]

What happened to the plus-minus sign that was in the original equation in the seventh question?

- Since there are no more square root terms, it is unnecessary.
- Each term with it was eventually squared, and a squared term must be positive it.
- Because the terms were moved from one side to another multiple times, it is no longer necessary.
- Any terms with it canceled out.

10.

Letting [math]b^2 = c^2 - a^2[/math], the equation no becomes:

[math]b^2 x^2 - y^2 = b^2 a^2[/math]

[math]x^2 / a^2 - y^2 / b^2 = 1[/math]

This is the equation of a hyperbola, centered at the origin with foci [math](-c,0)[/math] and [math](c,0)[/math]. Where does the substitution [math]b^2 = c^2 - a^2[/math] come from?

[math]b^2 x^2 - y^2 = b^2 a^2[/math]

[math]x^2 / a^2 - y^2 / b^2 = 1[/math]

This is the equation of a hyperbola, centered at the origin with foci [math](-c,0)[/math] and [math](c,0)[/math]. Where does the substitution [math]b^2 = c^2 - a^2[/math] come from?

- Since [math]b[/math] usually appears in the equation for a hyperbola, it must be included. Using the Pythagorean theorem, [math]a^2 +b^2 = c^2[/math], simply rearrange the equation.
- It has to be done, to ensure the asymptotes are related to the equation. The equations of the asymptotes are [math]y= pm b/a[/math], and knowing that [math]|c| > |a|[/math], squaring and rearranging results in [math]b^2 = c^2 - a^2[/math].
- It's done to simply the equation. [math]b[/math] is not defined yet, and since [math]|c| > |a|[/math], [math]c^2 > a^2[/math], and so there must be a positive number, [math]b^2[/math] such that [math]b^2 = c^2 - a^2[/math].
- Knowing that [math]b[/math] is the length of the semi-minor axis, a right triangle can be formed with the center of the hyperbola and either foci, with [math]b[/math] as the length of one leg of this triangle. Applying the Pythagorean theorem results in [math]b^2 + c^2 = a^2[/math], and simply rearrange.

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