Animated Solution for Mathematics - Differentiation: The normal to the curve, $x^2+2xy-3y^2=0$, at $(1,1)$

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden architecture of a curve. We are given the equation $x^2+2xy-3y^2=0$.

At first glance, it looks like a standard second-degree equation, but let us pause and look deeper. This is a homogeneous equation of degree $2$. In the world of coordinate geometry, this is a signal—a secret code—that this equation represents a pair of straight lines passing through the origin.

By factoring this, we reveal the truth:

This simplifies beautifully to $(x-y)(x+3y)=0$. We are dealing with two lines: $y=x$ and $y=-\frac{x}{3}$.

A Perpendicular Dance

We are focused on the point $P(1,1)$. Notice that $1-1=0$, so our point $P$ sits perfectly on the line $y=x$. Now, we need to find the normal at this point.

To find the normal, we must first find the tangent. We use implicit differentiation on the original equation:

Applying the power rule and the product rule, we get:

I know that differentiation can feel like a mechanical chore, but imagine the slope as the 'steepness' of the curve at that exact moment. When we substitute $x=1$ and $y=1$ into our derivative, we get:

Simplifying this, we find $4-4\frac{dy}{dx}=0$, which gives us the slope of the tangent $m_T=1$. The normal, being the loyal perpendicular partner, must have a slope $m_N=-\frac{1}{m_T}=-1$.

Using the point-slope form, the equation of our normal is $y-1=-1(x-1)$, which simplifies to the elegant linear equation $y=2-x$.

Now, the climax of our journey: where does this normal line meet the curve again? We substitute $y=2-x$ back into our original curve equation $x^2+2xy-3y^2=0$.

This gives us:

Expanding this, we get $x^2+4x-2x^2-3(4-4x+x^2)=0$. Combining the terms, we arrive at $-4x^2+16x-12=0$. Dividing by $-4$, we are left with the simple quadratic:

Factoring this, we find $(x-1)(x-3)=0$. The root $x=1$ is our starting point $P$. The root $x=3$ is our destination!

Plugging $x=3$ into our normal equation $y=2-x$, we get $y=2-3=-1$. Thus, the normal meets the curve again at the point $Q(3,-1)$.

We have arrived at the point $(3,-1)$. With a positive $x$-coordinate and a negative $y$-coordinate, we find ourselves firmly in the Fourth Quadrant.

You have successfully navigated the algebra, respected the geometry, and emerged with the answer. Remember, every equation is a story—you just have to learn how to read it.