Animated Solution for Mathematics - Conic Sections: Let the product of the focal distances of the point $(\sqrt{3},\frac{1}{2})$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

Imagine you are standing in a vast, silent observatory, looking at the geometry of the heavens. An ellipse is not just a shape; it is a path, a trajectory, a fundamental truth of planetary motion.

Today, we are going to unravel the secrets of a specific ellipse, defined by a point $P(\sqrt{3},\frac{1}{2})$ and a mysterious property regarding its foci. This is a detective story where we use the laws of geometry to uncover the hidden parameters of our system.

The Geometry of Focal Distances

Every point on an ellipse has a unique relationship with its two foci, $S$ and $S^{\prime }$. The distances from any point $P(x_1,y_1)$ to these foci are the focal distances, $r_1$ and $r_2$.

There is a beautiful, elegant property that links these distances to the ellipse's parameters: $r_1=a+e{x}_1$ and $r_2=a-e{x}_1$. When we multiply these, we get the product:

We are told this product is $\frac{7}{4}$, and we know our point $P$ has an $x$-coordinate of $\sqrt{3}$. Substituting these, we get:

This is our first key equation, a bridge between the size of the ellipse ($a$) and its shape ($e$).

The Algebraic Bridge

Now, we must ground our ellipse in the coordinate plane. Since point $P(\sqrt{3},\frac{1}{2})$ lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, its coordinates must satisfy the equation.

Plugging in our values, we get:

We recall the fundamental identity $b^2=a^2(1-e^2)$. By substituting this into our equation, we transform it into:

Clear the denominators by multiplying by $4a^2(1-e^2)$, and we arrive at:

The Quadratic Dance

We take our expression for $a^2$ from Phase 1, $a^2=\frac{7}{4}+3e^2$, and substitute it into our equation from Phase 2. The algebra here is a delicate dance:

The $4$ outside the bracket cancels the denominator in $\frac{7}{4}$, leaving us with $(7+12e^2)(1-e^2)$. Expanding this, we get $7-7e^2+12e^2-12e^4$, which simplifies to $7+5e^2-12e^4$.

Bringing everything to one side, we form the quadratic equation:

Solving this quadratic in $e^2$ by splitting the middle term, we find $(4e^2-3)(3e^2-2)=0$. This gives us two possible values for $e^2$: $\frac{3}{4}$ and $\frac{2}{3}$.

We have found our two eccentricities: $e_1=\frac{\sqrt{3}}{2}$ and $e_2=\frac{\sqrt{2}}{\sqrt{3}}$. The problem asks for the absolute difference between these two values.

We calculate:

Finding a common denominator of $2\sqrt{3}$, we get:

And there it is! The absolute difference is $\frac{3-2\sqrt{2}}{2\sqrt{3}}$. We have navigated the geometry, mastered the algebra, and arrived at the solution.