Q.Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 7, 0)$, $e = \frac{4}{3}$.

Concept: Hyperbola Standard Form (horizontal transverse axis).

Since the vertices are $(\pm 7, 0)$, the transverse axis is along the $x$-axis, centre at $(0,0)$, and $a = 7$.

Eccentricity $e = \frac{c}{a} = \frac{4}{3}$, so

$c = a \cdot e = 7 \cdot \frac{4}{3} = \frac{28}{3}$.

For a hyperbola, $c^2 = a^2 + b^2$. Thus

$b^2 = c^2 - a^2 = \left(\frac{28}{3}\right)^2 - 7^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}$. …

Why This Approach Works

When a hyperbola has vertices at $(\pm 7, 0)$, the center is at the origin and the transverse axis lies along the x-axis. This tells us the standard form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a$ is the distance from center to each vertex. The eccentricity $e = \frac{4}{3}$ relates $a$, $b$, and $c$ (the distance from center to each focus) through $c = ae$ and $c^2 = a^2 + b^2$.

The key insight: once we know $a$ and $e$, we can find $c$, then $b^2$, and the equation follows directly.

Step-by-Step Solution

1. Identify $a$ from the vertices.

   The vertices are $(\pm 7, 0)$, so the distance from the center $(0,0)$ to either vertex is $a = 7$.

2. Find $c$ using eccentricity.

   For any hyperbola, $e = \frac{c}{a}$. Given $e = \frac{4}{3}$ and $a = 7$:

3. Use the relationship between $a$, $b$, and $c$. For a hyperbola with horizontal transverse axis:

Substitute $c = \frac{28}{3}$ and $a = 7$:

4. Solve for $b^2$. Convert $49$ to ninths: $49 = \frac{441}{9}$. Then: …