Animated Solution for Mathematics - Conic Sections: The equations of two sides of a variable triangle are $x=0$ and $y=3$, and its third side is a tangent to the parabola $y^2=6x$. The locus of its circumcentre is :

Unlocking the Locus

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a problem; we are witnessing a dance between a parabola and a variable triangle.

Imagine standing on a coordinate plane. You see a fixed parabola, $y^2=6x$, standing still like a mountain. You see two fixed walls, $x=0$ and $y=3$, forming a perfect corner.

And then, there is a third line—a tangent—that slides along the parabola, constantly changing the shape of the triangle it forms with those walls. Our goal is to trace the path of the circumcenter of this shifting triangle.

The Foundation

First, we must understand our landscape. The parabola is given by $y^2=6x$.

Comparing this to the standard form $y^2=4ax$, we immediately identify $4a=6$, which gives us $a=\frac{3}{2}$. This value is the heartbeat of our parabola.

Now, look at the fixed lines. The line $x=0$ is the y-axis, and $y=3$ is a horizontal line. They meet at the point $A(0,3)$.

This is our anchor. Because these lines are perpendicular, any triangle formed by them and a third line will be a right-angled triangle at vertex $A$. This is a massive gift from geometry!

The Variable Tangent

The third side of our triangle is a tangent to the parabola. Using the slope form of a tangent, we write its equation as $y=mx+\frac{a}{m}$.

Substituting our value of $a$, we get:

This line is the 'variable' in our story. As $m$ changes, the line slides, and the triangle morphs. We need to find where this line hits our fixed walls.

For vertex $B$, we set $x=0$ in the tangent equation, yielding $y=\frac{3}{2m}$. Thus, $B=(0,\frac{3}{2m})$.

For vertex $C$, we set $y=3$ in the tangent equation: $3=mx+\frac{3}{2m}$. Solving for $x$, we find $x=\frac{6m-3}{2m^2}$. Thus, $C=(\frac{6m-3}{2m^2},3)$.

The Circumcenter's Secret

Here is where the magic happens. We know that for a right-angled triangle, the circumcenter $(h,k)$ is the midpoint of the hypotenuse $BC$.

We apply the midpoint formula:

We have our coordinates $(h,k)$ in terms of $m$. Now, we must eliminate $m$ to find the locus.

From the equation for $k$, we have $4mk=3+6m$, which rearranges to $m(4k-6)=3$, or:

The Final Synthesis

We substitute this into our expression for $h$. We rewrite $4h=\frac{6m-3}{m^2}$ as $4h=\frac{6}{m}-\frac{3}{m^2}$.

Substituting $\frac{1}{m}=\frac{2(2k-3)}{3}$, the algebra unfolds beautifully. After careful simplification—multiplying by 3 and dividing by 4—we arrive at:

Expanding this, we get $3h=-4k^2+18k-18$. Rearranging, we find the final equation:

Replacing $(h,k)$ with $(x,y)$, we obtain the locus:

$$4y^2-18y+3x+18=0$$

You have done it! You have tamed the variable triangle and found the path of its circumcenter. Remember, in JEE Advanced, the math is just the tool; the true skill is in visualizing the geometry before you even touch the pen.