Animated Solution for Mathematics - Differentiation: If the line $ax+by+c=0$ is a normal to the curve $xy=1$, then

Welcome, fellow traveler, to the fascinating world of coordinate geometry. Today, we are going to unravel the mystery behind the normal to a rectangular hyperbola.

Imagine you are standing on the curve $xy=1$. This curve is elegant, symmetric, and full of secrets. We are given a line $ax+by+c=0$ and told it is a normal to this curve.

A normal is simply a line perpendicular to the tangent at the point of contact. To solve this, we need to bridge the gap between calculus and algebra.

Our first mission is to find the slope of the tangent at any point $P(x^{\prime },y^{\prime })$ on our curve. We start with the equation $y=\frac{1}{x}$.

Using the power rule of differentiation, we find the derivative:

By substituting our point $P(x^{\prime },y^{\prime })$, we get the slope of the tangent, $m_T=-\frac{1}{x^{\prime 2}}$. This is the instantaneous rate of change at point $P$, telling us exactly how the curve is tilting at that precise moment.

Now, we need the slope of the normal. Geometry gives us a beautiful rule: the normal is perpendicular to the tangent.

If two lines are perpendicular, the product of their slopes is $-1$. So, $m_N=-\frac{1}{m_T}$.

Substituting our $m_T$, we get:

The negatives cancel out, leaving us with a clean, positive expression for the slope of the normal.

Next, let's look at the line $ax+by+c=0$. To find its slope, we rearrange it into the slope-intercept form $y=mx+C$.

By isolating $y$, we get $by=-ax-c$, which leads to:

The slope of our given line is $m_L=-\frac{a}{b}$. Since this line is the normal, its slope must equal the slope we derived from calculus. Thus, we equate them: $x^{\prime 2}=-\frac{a}{b}$.

Here is where we must be sharp. Look at the left side: $x^{\prime 2}$. Because $P$ is on the hyperbola $xy=1$, $x^{\prime }$ cannot be zero.

Therefore, $x^{\prime 2}$ is always strictly positive. This forces the right side, $-\frac{a}{b}$, to be strictly positive as well.

So, $-\frac{a}{b}>0$. Multiplying by $-1$ flips the inequality, giving us:

A fraction is negative only if the numerator and denominator have opposite signs. This means either $a>0$ and $b<0$, or $a<0$ and $b>0$.

We have successfully decoded the geometric condition! Keep this logic in your toolkit; it is the key to mastering many such problems in JEE Advanced.