Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $x=\frac{m}{n}$ ($m,n$ are co-prime natural numbers) be a solution of the equation $\cos (2{\sin }^{-1}x)=\frac{1}{9}$ and let $\alpha ,\beta (\alpha >\beta )$ be the roots of the equation $m{x}^2-nx-m+n=0$. Then the point $(\alpha ,\beta )$ lies on the line

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a mathematical investigation.

In the JEE Advanced arena, problems are rarely just about calculation; they are about connecting disparate worlds—trigonometry, algebra, and coordinate geometry—into a single, coherent narrative. Let us break down this problem, not as a series of steps, but as a journey.

The Trigonometric Mystery

We begin with the equation:

At first glance, this looks intimidating. We have an inverse trigonometric function nested inside a cosine function.

But here is the secret: whenever you see an inverse function, treat it as an angle. Let us define $\theta ={\sin }^{-1}x$. By the very definition of inverse functions, this implies that $\sin \theta =x$.

Now, our equation transforms into something much more elegant:

This is where your toolkit comes into play. We need to bridge the gap between $\cos (2\theta )$ and $\sin \theta$. We know the double-angle identity:

This is the perfect bridge! By substituting our definition, we get:

Now, the trigonometry has vanished, leaving us with a simple algebraic equation. Rearranging this, we find:

This simplifies beautifully to:

The Quadratic Bridge

We have arrived at $x=\pm \frac{2}{3}$. But wait—the problem gives us a crucial constraint: $x=\frac{m}{n}$, where $m$ and $n$ are co-prime natural numbers.

Since natural numbers are positive, we must reject the negative root. We are left with $x=\frac{2}{3}$, which gives us $m=2$ and $n=3$.

These are our keys to the next room of the puzzle. We are given the quadratic equation $m{x}^2-nx-m+n=0$.

Substituting our values, we get:

This simplifies to $2x^2-3x+1=0$. This is a classic quadratic. We can factorize this by splitting the middle term:

Grouping these terms, we get $2x(x-1)-1(x-1)=0$, leading us to $(2x-1)(x-1)=0$. The roots are $x=1$ and $x=\frac{1}{2}$.

The Coordinate Geometry Finale

We are told that $\alpha$ and $\beta$ are the roots, with the condition $\alpha >\beta$. This is not just a detail; it is a directive.

Since $1>\frac{1}{2}$, we must assign $\alpha =1$ and $\beta =\frac{1}{2}$. Our point is $(1,\frac{1}{2})$.

Now, we test our options. We need to see which line passes through $(1,\frac{1}{2})$. Let us test the fourth option: $5x+8y=9$.

Substituting our coordinates, we get:

It fits perfectly!

This problem was a test of your ability to maintain composure. We started with a trigonometric expression, moved through an algebraic constraint, solved a quadratic, and finished with coordinate geometry.

The key to success in JEE Advanced is not just knowing the formulas, but knowing how to weave them together. You have successfully navigated the connections.

Keep this mindset—see the connections, respect the constraints, and the solution will always reveal itself.