Animated Solution for Mathematics - Straight Lines: If the pair of lines $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ intersect on the y-axis then

Unlocking the Y-Axis Intersection

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, looks like a daunting algebraic beast.

We are staring at the general second-degree equation:

It is a classic, a cornerstone of coordinate geometry. Today, we aren't just solving it; we are going to understand the geometry breathing beneath the algebra.

The General Condition

Before we dive into the specifics of the y-axis, we must respect the structure of the equation. For this expression to represent a pair of straight lines, it must satisfy a very strict condition.

We call this the vanishing of the discriminant, denoted by $\Delta$. The condition is:

Think of this as the 'existence' condition. If this is not zero, the equation doesn't represent lines at all—it might be a hyperbola, an ellipse, or something else entirely. We keep this equation in our back pocket, like a master key, ready to unlock the final answer later.

The Geometric Insight

The problem gives us a beautiful, simplifying constraint: the lines intersect on the y-axis. Now, pause for a moment.

Close your eyes and visualize the Cartesian plane. If a point lies on the y-axis, what is the one thing we know for certain? Its x-coordinate is zero.

This is not just a coordinate; it is a mathematical lever. By setting $x=0$, we are essentially 'slicing' the geometry to see what remains at the y-axis.

The Quadratic Pivot

Let us perform the substitution. When we plug $x=0$ into our general equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, the terms $a{x}^2$, $2hxy$, and $2gx$ vanish instantly.

They become zero. We are left with a simple, elegant quadratic equation in terms of $y$:

This is where the magic happens. This quadratic equation tells us the y-coordinates of the points where our lines cross the y-axis.

But wait—the problem tells us the lines intersect at a point. Singular. One point. If this quadratic had two distinct roots, it would mean the lines intersect the y-axis at two different places.

That contradicts our premise! Therefore, for the lines to intersect at exactly one point, this quadratic must have equal roots. In the language of algebra, the discriminant of this quadratic must be zero.

The Elegant Cancellation

Let us apply the discriminant condition $D=B^2-4AC=0$ to our quadratic $b{y}^2+2fy+c=0$. Here, $A=b$, $B=2f$, and $C=c$.

So, $(2f{)}^2-4(b)(c)=0$. This simplifies beautifully to $4f^2-4bc=0$, or simply:

Now, we bring back our master key, the $\Delta =0$ condition:

We also know that $f^2=bc$. If we multiply both sides of our new relation by $a$, we get $a{f}^2=abc$.

Look at the $\Delta$ equation again. We can replace $abc$ with $a{f}^2$. Substituting this in, we get:

The terms $a{f}^2$ and $-a{f}^2$ cancel out perfectly, leaving us with $2fgh-b{g}^2-c{h}^2=0$. Rearranging this, we arrive at the final, elegant result:

Look at what we have achieved. We started with a complex, intimidating general equation, and through the power of geometric visualization and algebraic substitution, we distilled it into a clean, symmetric relationship.

This is the essence of JEE Advanced physics and mathematics: it is not about memorizing formulas; it is about seeing the geometry, applying the right constraints, and letting the algebra reveal the truth. You have the tools. You have the logic. Now, go forth and conquer the next problem with this same clarity.