Animated Solution for Mathematics - Conic Sections: The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is

Welcome, fellow traveler in the realm of coordinate geometry. Today, we are not just solving a problem; we are witnessing a beautiful conversation between two fundamental shapes: the ellipse and the hyperbola.

These curves, born from the intersection of a plane and a cone, share a deep, mathematical bond. When we are told that their foci coincide, we are being given a secret key to unlock the hidden parameters of the ellipse.

Consider the equation of the hyperbola:

At first glance, the right-hand side is not $1$. To transform this into the standard form, we multiply the entire equation by $25$:

To identify the parameters $a_h^2$ and $b_h^2$, we rewrite the coefficients in the denominators:

By comparing this to the standard form $\frac{x^2}{a_h^2}-\frac{y^2}{b_h^2}=1$, we find $a_h^2=\frac{144}{25}$ and $b_h^2=\frac{81}{25}$. Taking the square roots, we obtain $a_h=\frac{12}{5}$ and $b_h=\frac{9}{5}$.

The eccentricity $e_h$ of the hyperbola is defined by the formula $e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}$. Substituting our values:

Simplifying the fraction $\frac{81}{144}$ by dividing both terms by $9$, we get $\frac{9}{16}$. Thus:

The foci of a standard hyperbola lie at $(\pm a_h{e}_h,0)$. Calculating the product $a_h{e}_h$:

Therefore, the foci of the hyperbola are located at $(\pm 3,0)$.

We now turn to the ellipse: $\frac{x^2}{16}+\frac{y^2}{b^2}=1$. Since its foci coincide with the hyperbola, the foci are also at $(\pm 3,0)$.

Comparing this to the standard ellipse $\frac{x^2}{a_e^2}+\frac{y^2}{b_e^2}=1$, we identify $a_e^2=16$, which implies $a_e=4$. The focal distance for an ellipse is given by $a_e{e}_e=3$.

Substituting $a_e=4$ into the focal distance equation, we find $4e_e=3$, or $e_e=\frac{3}{4}$. This confirms we are dealing with an ellipse, as $e_e<1$.

We use the fundamental relationship for an ellipse, $b^2=a_e^2(1-e_e^2)$, to find the unknown parameter $b^2$. Substituting $a_e^2=16$ and $e_e=\frac{3}{4}$:

The final result is $b^2=7$.