Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.1 Exercises - Page 692: 48

$$\frac{{{x^2}}}{{256}} - \frac{{{y^2}}}{{144}} = 1$$

$$\eqalign{ & {\text{Focus }}\left( {20,0} \right) \cr & {\text{Asymptotes }}y = \pm \frac{3}{4}x \cr & {\text{Find the center of the hyperbola}} \cr & {\text{let }}y = y \cr & \frac{3}{4}x = - \frac{3}{4}x \cr & \frac{3}{4}x + \frac{3}{4}x = 0 \cr & x = 0 \to y = 0 \cr & {\text{The center of the hyperbola is }}\left( {0,0} \right) = \left( {h,k} \right),{\text{ so}} \cr & {\text{Center }}\left( {0,0} \right){\text{ Focus }}\left( {20,0} \right) \cr & {\text{The standard equation of the hyperbola is}} \cr & \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr & {\text{Foci }}\left( { \pm c,0} \right) \to c = 20 \cr & {\text{Asymptotes }}y = \pm \frac{b}{a}x \cr & \frac{b}{a} = \frac{3}{4} \to b = \frac{3}{4}a \cr & {\text{We know that }}{c^2} = {a^2} + {b^2} \cr & {\left( {20} \right)^2} = {a^2} + {\left( {\frac{3}{4}a} \right)^2} \cr & {\left( {20} \right)^2} = {a^2} + \frac{9}{{16}}{a^2} \cr & \frac{{25}}{{16}}{a^2} = {\left( {20} \right)^2} \cr & {a^2} = \frac{{16}}{{25}}{\left( {20} \right)^2} \cr & {a^2} = 256 \cr & b = \frac{3}{4}a = \frac{3}{4}\left( {16} \right) = 12 \cr & {b^2} = 144 \cr & \cr & \underbrace {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1}_ \downarrow \cr & \frac{{{x^2}}}{{256}} - \frac{{{y^2}}}{{144}} = 1 \cr} $$