Graphing Equation of Hyperbola Problem

a) Write the given equation in standard form by dividing both sides by 144:

\[ \frac{9x^2}{144} - \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} - \frac{y^2}{9} = 1 \implies \frac{x^2}{4^2} - \frac{y^2}{3^2} = 1 \]

Comparing this with standard equation (I), we find \( a = 4 \) and \( b = 3 \).

• x-intercepts: Set \( y = 0 \): \[ \frac{x^2}{4^2} = 1 \implies x^2 = 16 \implies x = \pm 4 \]
• y-intercepts: Set \( x = 0 \): \[ \frac{-y^2}{9} = 1 \implies y^2 = -9 \] No real solution, hence no y-intercepts.

b) Calculate \( c \) using the relation \( c^2 = a^2 + b^2 \):

\[ c^2 = 4^2 + 3^2 = 16 + 9 = 25 \implies c = \pm 5 \]

The coordinates of the foci are \( F_1 = (5, 0) \) and \( F_2 = (-5, 0) \).

c) Sketching the graph:

1. Draw the asymptotes: \( y = -\frac{3}{4}x \) and \( y = \frac{3}{4}x \).
2. Plot the x-intercepts at \( (4,0) \) and \( (-4,0) \).
3. Find auxiliary points: let \( x = 6 \): \[ 9(6)^2 - 16y^2 = 144 \implies 324 - 144 = 16y^2 \implies y^2 = \frac{180}{16} = \frac{45}{4} \implies y = \pm \frac{3\sqrt{5}}{2} \] Points \( \left(6, \frac{3\sqrt{5}}{2}\right) \), \( \left(6, -\frac{3\sqrt{5}}{2}\right) \), and their symmetric counterparts lie on the graph.

a) Intercepts:

• Divide by 9 to get standard form: \(\dfrac{x^2}{3^2} - \dfrac{y^2}{3^2} = 1\), where \( a = 3, b = 3 \).
• x-intercepts: Set \( y = 0 \implies x^2 = 9 \implies x = \pm 3 \).
• y-intercepts: Set \( x = 0 \implies -y^2 = 9 \implies y^2 = -9 \) (No real solution, so no y-intercepts).

b) Foci:

\[ c^2 = a^2 + b^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies c = \pm\sqrt{18} = \pm 3\sqrt{2} \]

The foci are \( (3\sqrt{2}, 0) \) and \( (-3\sqrt{2}, 0) \).

c) Sketching details: Asymptotes are \( y = -x \) and \( y = x \), passing through the origin with vertices at \( (\pm 3, 0) \).