Animated Solution for Mathematics - Conic Sections: An ellipse has $OB$ as semi minor axis, $F$ and $F^{\prime }$ its focii and the angle $FB{F}^{\prime }$ is a right angle. Then the eccentricity of the ellipse is

Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a geometric masterpiece. We are investigating the relationship between the foci and the minor axis of an ellipse.

Imagine you are standing at the center of an ellipse, $O(0,0)$. You look out toward the minor axis, where the point $B(0,b)$ sits, and then you look toward the foci $F(ae,0)$ and $F^{\prime }(-ae,0)$. The problem states that the angle $\angle FB{F}^{\prime }$ is a perfect $90^{\circ }$.

First, let us ground ourselves in the definition of an ellipse. The sum of the distances from any point on the ellipse to its two foci is a constant, $2a$. This is the heartbeat of the ellipse.

When we look at point $B$, which lies on the minor axis, we see a beautiful symmetry. Because the ellipse is symmetric about the $y$-axis, the distance from $B$ to $F$ must be identical to the distance from $B$ to $F^{\prime }$.

Mathematically, we state this as:

Since $BF=B{F}^{\prime }$, we immediately find that $2BF=2a$, which simplifies to $BF=a$. This is a profound realization! The distance from the endpoint of the minor axis to either focus is exactly equal to the semi-major axis $a$.

Now, we turn our attention to the triangle $△FB{F}^{\prime }$. We are given that $\angle FB{F}^{\prime }=90^{\circ }$. This is our golden key.

In any right-angled triangle, the relationship between the sides is governed by the eternal truth of Pythagoras. Here, the hypotenuse is the segment connecting the two foci, $F{F}^{\prime }$. The length of this segment is the distance between $(ae,0)$ and $(-ae,0)$, which is $2ae$.

Applying the Pythagorean theorem, we write:

Substituting our known values, $BF=a$, $B{F}^{\prime }=a$, and $F{F}^{\prime }=2ae$, we get:

Look at the equation above. It is simple, yet it holds the secret to the ellipse's eccentricity. Let us expand the left side:

Here is where the magic happens. Since $a$ represents the semi-major axis, we know $a eq 0$. We can safely divide both sides by $2a^2$:

Solving for $e^2$, we find:

Taking the square root, and remembering that eccentricity must be positive, we arrive at our final destination:

We have arrived at the answer, but notice the journey. We started with a simple geometric condition—a right angle—and through the properties of symmetry and the Pythagorean theorem, we uncovered the specific eccentricity required to create this shape.

This is the essence of JEE mathematics: taking a complex, abstract description and distilling it into a clear, logical sequence of steps. You have done well to follow this path. Keep this clarity of thought, and no problem will ever be too daunting.