Animated Solution for Mathematics - Conic Sections: On the ellipse $4x^2+9y^2=1$, the points at which the tangents are parallel to the line $8x=9y$ are

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving a coordinate geometry problem; we are uncovering the hidden symmetry of the ellipse.

Imagine yourself standing at the center of the ellipse defined by $4x^2+9y^2=1$. This shape is a perfect, flattened circle, a beautiful manifestation of conic sections.

Our mission is to find the exact coordinates where a tangent line, sliding along the perimeter, becomes perfectly parallel to the line $8x=9y$.

Before we dive into the calculus, we must see the ellipse in its standard form. The equation $4x^2+9y^2=1$ is a bit disguised. To reveal its true nature, we rewrite it as:

Here, we identify our semi-axes squared: $a^2=\frac{1}{4}$ and $b^2=\frac{1}{9}$. This tells us exactly how much the ellipse is stretched along the $x$ and $y$ axes.

Now, look at our target line: $8x=9y$. Rearranging this into the slope-intercept form $y=mx+c$, we get $y=\frac{8}{9}x$. Our target slope is $m=\frac{8}{9}$.

We are looking for the points on the ellipse where the derivative $\frac{dy}{dx}$ is exactly $\frac{8}{9}$.

While we could use implicit differentiation, the JEE rewards those who know their tools. For any ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the point of contact for a tangent with slope $m$ is given by the elegant coordinate pair:

I know this formula looks intimidating, but let's take a breath and look at its soul. The denominator $\sqrt{a^2{m}^2+b^2}$ is the key. It acts as a bridge between the slope of the line and the geometry of the ellipse.

Let's calculate it together. Substituting our values $a^2=\frac{1}{4}$, $b^2=\frac{1}{9}$, and $m=\frac{8}{9}$, we get:

Isn't that satisfying? The complexity collapses into a simple fraction, $\frac{5}{9}$.

Now, we simply plug this denominator back into our coordinate formulas. For the $x$-coordinate:

And for the $y$-coordinate:

We have found our points! Because of the symmetry of the ellipse, we have two points: $\left(-\frac{2}{5},\frac{1}{5}\right)$ and $\left(\frac{2}{5},-\frac{1}{5}\right)$.

Think about what we just did. We took a line, a curve, and a set of algebraic constraints, and we found the exact locations where they harmonize.

In the JEE, you will often find that the most complex-looking problems are just waiting for you to apply the right symmetry. Never fear the algebra; embrace the geometry.

You have the tools, you have the logic, and now, you have the result. Keep pushing, keep questioning, and keep falling in love with the physics and math that govern our universe.