Animated Solution for Mathematics - Conic Sections: For the Hyperbola $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$, which of the following remains constant when $\alpha$ varies?

To understand the hyperbola defined by the equation

we compare it to the standard form of a hyperbola, which is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

By inspection, we identify the parameters as $a^2={\cos }^2\alpha$ and $b^2={\sin }^2\alpha$. Consequently, the semi-axes are $a=\cos \alpha$ and $b=\sin \alpha$.

These parameters represent the "DNA" of the hyperbola, dictating its shape and the position of its vertices at $(\pm a,0)$. Since both $a$ and $b$ depend on $\alpha$, the vertices shift as the parameter varies.

Next, we investigate the eccentricity $e$, which measures the "openness" of the curve. We utilize the fundamental hyperbola relationship:

Substituting our identified values into this equation, we obtain:

Rearranging to solve for $e^2$, we find:

This simplifies to $e^2=1+{\tan }^2\alpha$. Applying the trigonometric identity $1+{\tan }^2\alpha ={\sec }^2\alpha$, we conclude that $e=\sec \alpha$. Because $e$ is a function of $\alpha$, the eccentricity is not constant.

Finally, we examine the coordinates of the foci, which are located at $(\pm ae,0)$. This is where the invariance is revealed.

Given $a=\cos \alpha$ and $e=\sec \alpha$, we calculate the product $ae$:

Since $\sec \alpha =\frac{1}{\cos \alpha }$, the expression becomes:

The parameter $\alpha$ has completely vanished from the expression. Therefore, the foci are fixed at the coordinates $(\pm 1,0)$.

No matter how the hyperbola morphs, its foci remain stubbornly and beautifully constant.