Animated Solution for Mathematics - Conic Sections: Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^2}{4}+\frac{y^2}{2}=1$ from any of its foci?

Unveiling the Auxiliary Circle

Welcome, fellow explorer of the mathematical universe! Today, we are going to peel back the layers of a classic JEE Advanced problem. It might look like a standard coordinate geometry question, but beneath the surface lies a beautiful, elegant property of ellipses that, once understood, turns a daunting calculation into a simple verification.

Imagine you are standing on the edge of an ellipse, watching a tangent line glide smoothly around its perimeter. From one of the foci, you drop a perpendicular line to this tangent. As the tangent moves, the point where your perpendicular meets the tangent traces a path. What is this path? Let us find out.

First, let us look at our ellipse:

By comparing this to the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we immediately see that $a^2=4$ and $b^2=2$. This tells us that our semi-major axis $a$ is $2$.

The ellipse is stretched along the $x$-axis, centered at the origin $(0,0)$. The foci, those two magical points that define the ellipse, lie on the major axis at $(\pm ae,0)$. While we could calculate the eccentricity $e$, the beauty of this problem is that we do not actually need to know the exact coordinates of the foci to find the locus.

Imagine drawing a tangent line at any point on this ellipse. Now, from one of the foci, you drop a perpendicular line to this tangent. This intersection point—the foot of the perpendicular—is what we are tracking.

If you were to attempt this by brute force, you would define the tangent line using a slope parameter $m$, write the equation of the perpendicular line, find their intersection point $(x,y)$, and then laboriously eliminate $m$ to find the relationship between $x$ and $y$. It is a path filled with algebraic pitfalls and potential for error.

Here is where we pause and appreciate the elegance of geometry. There is a standard, powerful result in the study of conics: the locus of the foot of the perpendicular drawn from any focus to any tangent of an ellipse is its auxiliary circle.

The auxiliary circle is defined as the circle concentric with the ellipse, with a radius equal to the semi-major axis $a$. This is not just a shortcut; it is a fundamental truth about how ellipses are constructed. The equation of this circle is simply $x^2+y^2=a^2$.

Since we determined $a^2=4$, our locus is the circle:

Now that we have our path, $x^2+y^2=4$, the problem becomes a simple test of membership. We are given four points and asked which one lies on this locus. Let us test the point $(-1,\sqrt{3})$:

It fits perfectly! The point $(-1,\sqrt{3})$ satisfies the equation of the auxiliary circle. This confirms that this point is indeed a valid position for the foot of the perpendicular.

We have successfully navigated the problem, not by drowning in algebra, but by leveraging the deep, geometric soul of the ellipse. Keep this property in your toolkit—it is a powerful ally for your JEE journey!