Animated Solution for Mathematics - Conic Sections: If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:

A Journey into Eccentricity

Welcome, fellow explorer of mathematics! Today, we are going to unravel the mysteries of the hyperbola. It is not just an equation on a page; it is a beautiful, sweeping curve that defines the paths of comets and the geometry of our universe.

Let us dive into this problem and see how we can master it.

Imagine you are looking at a standard hyperbola centered at the origin, opening along the $x$-axis. Its equation is:

The problem provides us with two vital clues: the length of the conjugate axis is $5$, and the distance between the foci is $13$.

Geometrically, the conjugate axis is the vertical segment that helps define the 'box' of the hyperbola. Its length is defined as $2b$. So, we immediately have the equation $2b=5$.

Dividing by $2$, we find that $b=\frac{5}{2}$. This is our first anchor point.

Next, let us look at the foci. The distance between the two foci is defined as $2ae$, where $e$ is the eccentricity. The problem tells us this distance is $13$.

Thus, $2ae=13$. Dividing by $2$, we get $ae=\frac{13}{2}$. This product $ae$ is a powerful shortcut that will simplify our algebra significantly.

Now, we need to bridge the gap between $a$, $b$, and $e$. The fundamental identity for any hyperbola is:

This equation is the heart of the hyperbola's definition. Let us expand the right side: $b^2=a^2{e}^2-a^2$.

Notice how $a^2{e}^2$ is simply the square of $ae$. We can rewrite this as:

This is a brilliant transformation! It allows us to substitute our known values directly without dealing with $e$ as an unknown variable just yet.

We have $b=\frac{5}{2}$ and $ae=\frac{13}{2}$. Let us plug these into our identity:

Calculating the squares, we get:

Now, we isolate $a^2$ by rearranging the terms: $a^2=\frac{169}{4}-\frac{25}{4}$.

Since the denominators are identical, this subtraction is straightforward: $a^2=\frac{144}{4}$, which simplifies beautifully to $a^2=36$. Taking the square root, we find $a=6$.

Remember, we only take the positive root because $a$ represents a physical length.

We are almost at the finish line. We know $ae=\frac{13}{2}$ and we have just discovered that $a=6$.

Substituting $a$ into our equation for the foci distance, we get $6e=\frac{13}{2}$. To isolate $e$, we divide both sides by $6$:

And there it is! The eccentricity of our hyperbola is $\frac{13}{12}$.

It is a value greater than $1$, which is exactly what we expect for a hyperbola. You have navigated the geometry, utilized the master identity, and performed the algebra with precision. Well done!