Animated Solution for Mathematics - Conic Sections: Let $A(\sec \theta ,2\tan \theta )$ and $B(\sec \varphi ,2\tan \varphi )$, where $\theta +\varphi =\pi /2$, be two points on the hyperbola $2x^2-y^2=2$. If $(\alpha ,\beta )$ is the point of the intersection of the normals to the hyperbola at A and B, then $(2\beta {)}^2$ is equal to .

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant dance of geometry and trigonometry on the canvas of a hyperbola.

When you look at the equation $2x^2-y^2=2$, I want you to see more than just variables and numbers. I want you to see a structure, a conic section that defines the path of celestial bodies and the limits of our physical world.

Let us begin by standardizing our battlefield. We divide the entire equation by $2$, transforming it into:

Immediately, the parameters reveal themselves: $a^2=1$ and $b^2=2$. This is our foundation. Without this clarity, we are lost in the woods; with it, we have a map.

The Normal Equation

We are dealing with normals—lines perpendicular to the tangent at a specific point. Many students panic when they see normals, thinking they must calculate the derivative, find the tangent slope, invert it, and then construct the line equation.

While that is a valid path, it is the path of the amateur. We are aiming for the elite level. We use the standard formula for the normal to a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at a point $(x_1,y_1)$, which is:

This formula is your best friend. It encapsulates the geometry of the hyperbola into a single, powerful algebraic tool.

Let us apply this to point $A(\sec \theta ,2\tan \theta )$. Substituting our values, we get:

With a bit of trigonometric simplification—recalling that $\frac{1}{\sec \theta }=\cos \theta$ and $\frac{1}{\tan \theta }=\cot \theta$—the equation simplifies beautifully to:

This is the equation of our first normal. It is clean, it is precise, and it is ready.

Now, we turn our attention to point $B(\sec \varphi ,2\tan \varphi )$. We could repeat the entire derivation, but why should we? Mathematics is the study of patterns.

Since the structure of point $B$ is identical to point $A$, the equation of the normal at $B$ must follow the same pattern:

Here is where the magic happens. The problem gives us the constraint $\theta +\varphi =\pi /2$. This is not a coincidence; it is a bridge.

It allows us to express $\varphi$ as $\pi /2-\theta$. When we substitute this into our second normal equation, the trigonometric identities come alive:

Suddenly, our second normal equation transforms into:

We have successfully unified our system under a single parameter, $\theta$.

We now stand before a system of two linear equations:

1) $x\cos \theta +y\cot \theta =3$ 2) $x\sin \theta +y\tan \theta =3$

Our goal is to find the intersection point $(\alpha ,\beta )$. The question asks for $(2\beta {)}^2$, which means we only care about $\beta$, the $y$-coordinate. To isolate $y$, we must eliminate $x$.

We make the coefficients of $x$ identical by multiplying the first equation by $\sin \theta$ and the second by $\cos \theta$.

Equation 1 becomes:

Since $\cot \theta =\frac{\cos \theta }{\sin \theta }$, the middle term simplifies to $y\cos \theta$. Thus:

Equation 2 becomes:

Since $\tan \theta =\frac{\sin \theta }{\cos \theta }$, the middle term simplifies to $y\sin \theta$. Thus:

Now, subtract the second from the first. The $x$ terms vanish into thin air, leaving us with:

This is the moment of truth. We factor out a negative sign on the right side:

Assuming $\cos \theta eq \sin \theta$, we divide both sides by $(\cos \theta -\sin \theta )$ to reveal the elegant result:

Thus, $\beta =-3$.

We have arrived. The question asks for $(2\beta {)}^2$. Substituting $\beta =-3$, we get:

Take a moment to appreciate this. We started with a complex hyperbola and two arbitrary points, and through the power of symmetry and trigonometric identities, we collapsed the complexity into a single, solid integer.

This is the beauty of JEE Advanced mathematics. It is not about brute force; it is about finding the path of least resistance through the logic. You have mastered this problem. Keep this confidence, keep this curiosity, and you will conquer any challenge that comes your way. The final answer is 36.