Animated Solution for Mathematics - Conic Sections: Find the co-ordinates of all the points $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, for which the area of the triangle $PON$ is a maximum, where $O$ denotes the origin and $N$, the foot of the perpendicular from $O$ to the tangent at $P$.

Imagine standing at the origin of a coordinate plane, watching a point $P$ trace the graceful curve of an ellipse defined by:

As $P$ moves, we draw a tangent line, and from our vantage point at the origin, we drop a perpendicular to that tangent. The point where this perpendicular meets the tangent is $N$. We are forming a triangle, $\Delta PON$, and our mission is to find the exact position of $P$ that makes this triangle as large as possible.

When we face an ellipse, the standard Cartesian coordinates $(x,y)$ can sometimes feel like a heavy anchor. To set our math free, we use the parametric form.

We define $P$ as $(a\cos \theta ,b\sin \theta )$, where $\theta$ is the eccentric angle. This single variable, $\theta$, now controls the entire system.

With $P$ defined, the tangent line at $P$ becomes a simple, elegant equation:

The distance $ON$ is the perpendicular distance from the origin $(0,0)$ to the tangent line. Using the standard distance formula, we find:

Next, we need the hypotenuse $OP$. Since $O$ is the origin:

Because $\Delta PON$ is right-angled at $N$, we invoke the Pythagorean theorem: $P{N}^2=O{P}^2-O{N}^2$. After simplifying the algebraic terms, we find:

AM-GM

The area $Z$ is given by $Z=\frac{1}{2}\times ON\times PN$. When we multiply our expressions for $ON$ and $PN$, the square roots in the denominator cancel out perfectly.

To maximize this, we divide the numerator and denominator by ${\cos }^2\theta$, transforming the expression into a function of $t=\tan \theta$. We are left with a denominator of the form:

The Arithmetic Mean-Geometric Mean (AM-GM) inequality tells us that for positive terms, the sum is minimized when the terms are equal. We set $\frac{b^2}{t}=a^{2}t$, which leads us to $t^2=\frac{b^2}{a^2}$, or $\tan \theta =\pm \frac{b}{a}$.

Substituting these values back into our parametric coordinates, we find the final points $P$ are:

You have navigated the complexity and found the symmetry. This is the essence of JEE Advanced mathematics: finding the simple, elegant truth hidden beneath the surface of a complex problem.