Animated Solution for Mathematics - Conic Sections: If the line $3x-2y+12=0$ intersects the parabola $4y=3x^2$ at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

A Tale of Two Curves

Imagine you are standing on a coordinate plane, looking at a beautiful, upward-opening parabola defined by the equation $4y=3x^2$. Its vertex rests perfectly at the origin $O(0,0)$.

Now, imagine a straight line, $3x-2y+12=0$, slicing through this parabola like a blade, creating two distinct intersection points, $A$ and $B$. Your mission is to find the angle $\theta$ subtended by the segment $AB$ at the vertex $O$.

This is not just a problem; it is a journey into the elegance of coordinate geometry.

Many students, upon seeing this, immediately reach for their pens to solve the system of equations. They try to substitute $y=\frac{3x^2}{4}$ into the line equation.

While this is a valid path, it is a treacherous one. You would end up with a quadratic equation in $x$, and the roots would likely be messy, irrational numbers.

Then, you would have to find the corresponding $y$-coordinates, calculate the slopes $m_1$ and $m_2$ of lines $OA$ and $OB$, and finally use the angle formula. It is a path filled with potential for arithmetic slips.

Homogenization

We want the angle between the lines $OA$ and $OB$. These lines pass through the origin.

If we could find the joint equation of these two lines, we would have the slopes $m_1$ and $m_2$ almost instantly. This is where the technique of Homogenization comes into play.

The core idea is to make the equation of the parabola homogeneous of degree 2 by using the linear equation of the line.

First, let us rewrite our line equation $3x-2y+12=0$ to isolate the constant: $3x-2y=-12$. Dividing by $-12$, we get:

This expression is our 'magic key'.

Now, look at the parabola: $3x^2=4y$. The left side is degree 2, but the right side is degree 1. To balance this, we multiply the right side by our magic key, which is equal to 1:

Watch how the algebra unfolds. The $4$ and $12$ simplify to a $3$ in the denominator. Cross-multiplying that $3$ gives us $9x^2=y(2y-3x)$.

Expanding this, we get $9x^2=2y^2-3xy$. Rearranging everything to one side, we arrive at the beautiful, homogeneous second-degree equation:

This equation represents the pair of lines $OA$ and $OB$. It is a masterpiece of algebraic balance.

To find the slopes, we divide the entire equation by $x^2$, assuming $x eq 0$. Let $m=\frac{y}{x}$ be the slope of the lines.

The equation transforms into a quadratic in $m$:

Factoring this quadratic is straightforward: $(2m+3)(m-3)=0$. This gives us two distinct slopes: $m_1=3$ and $m_2=-\frac{3}{2}$.

These are the slopes of our lines $OA$ and $OB$.

Finally, we use the standard formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

Substituting our values:

Thus, the final answer is:

We have arrived at the answer, not through brute force, but through the sheer elegance of mathematical structure. Remember, in JEE Advanced, it is often not about how hard you work, but how cleverly you use the tools in your arsenal.