Animated Solution for Mathematics - Conic Sections: The normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{9}=1$ at the point $(8,3\sqrt{3})$ on it passes through the point :

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey to understand the elegant, sweeping curves of a hyperbola.

When you look at the equation $\frac{x^2}{a^2}-\frac{y^2}{9}=1$, do not just see a collection of variables. See a challenge. We are given a point $P(8,3\sqrt{3})$ that sits proudly on this curve, and our mission is to find the normal line passing through it.

In mathematics, when you are missing a piece of the puzzle, you must look for clues in the environment. We are told that the point $P(8,3\sqrt{3})$ lies on the hyperbola, which means it must satisfy the curve's equation.

Let us substitute our coordinates into the equation:

Take a deep breath and perform the arithmetic. Since $8^2=64$ and $(3\sqrt{3}{)}^2=9\times 3=27$, the equation becomes:

See how the complexity melts away? Since $\frac{27}{9}=3$, the equation simplifies to $\frac{64}{a^2}-3=1$, or $\frac{64}{a^2}=4$. With a simple rearrangement, we find that $a^2=16$.

We have successfully completed our hyperbola: $\frac{x^2}{16}-\frac{y^2}{9}=1$.

Now, we need the normal. In the JEE Advanced arena, we value efficiency, so we use the standard formula for the normal to a hyperbola at a point $(x_1,y_1)$:

We know $a^2=16$, $b^2=9$, and our point $P$ is $(8,3\sqrt{3})$. Let us plug these values into our formula:

Now, let us simplify the expression. The first term, $\frac{16x}{8}$, is $2x$.

The second term, $\frac{9y}{3\sqrt{3}}$, simplifies to $\frac{3y}{\sqrt{3}}$, which is $\sqrt{3}y$. On the right side, $16+9=25$.

Our normal line equation is:

To verify, we test the point $(-1,9\sqrt{3})$:

It matches perfectly! The left-hand side equals the right-hand side. We have arrived at our destination.