Nonlinear Equations and Functions Question

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For the hyperbola centered at the origin with foci

$F_1 = (-c,0)$ and

$F_2=(c,0)$ , derive the equation of the hyperbola.

Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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?

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.
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 $b^2 = c^2 - a^2$ .
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 $b^2 = c^2 - a^2$ .
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.

Question Info

Nonlinear Equations and Functions Question

Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3