Animated Solution for Mathematics - Conic Sections: Comprehension Passage

A Journey into Tangency

Imagine you are standing on the Cartesian plane, looking at a region $R$ defined by $x\ge 0$ and $y^2\le 4-x$. On your left, the y-axis stands like an impenetrable wall.

To your right, a beautiful parabola, $y^2=4-x$, curves gracefully, opening towards the left. The region $R$ is the space trapped between this wall and the curve.

We want to place the largest possible circle inside this region. This is a classic JEE Advanced challenge that tests your ability to bridge the gap between geometric intuition and algebraic rigor.

To maximize the radius of a circle contained within this region, we must push it as far left as possible until it hits the y-axis. Since the circle must touch the y-axis at the origin to be as large as possible, its center must lie on the x-axis at some point $(r,0)$, where $r$ is the radius.

This gives us our first powerful insight: the circle's equation is $(x-r{)}^2+y^2=r^2$. Expanding this, we get $x^2-2rx+r^2+y^2=r^2$, which simplifies to:

This is the equation of our family of circles $F$.

As we increase the radius $r$, the circle grows. Eventually, it will hit the parabola $y^2=4-x$. If it grows any further, it will cross the boundary, which is not allowed.

Therefore, the largest circle must be tangent to the parabola. This is the crucial moment in our journey.

Tangency means the circle and the parabola share exactly one point of contact. Algebraically, this means that if we substitute the parabola's equation into the circle's equation, the resulting quadratic equation in $x$ must have a repeated root.

Let's perform the substitution. We know $y^2=4-x$. Plugging this into our circle equation $x^2+y^2-2rx=0$, we get:

Rearranging the terms, we obtain the quadratic equation:

For this circle to be tangent to the parabola, this quadratic must have a discriminant $\Delta =0$. The discriminant is given by $\Delta =b^2-4ac$.

Here, $a=1$, $b=-(2r+1)$, and $c=4$. Thus:

This simplifies to $(2r+1{)}^2-16=0$, or $(2r+1{)}^2=16$.

Taking the square root of both sides, we get $2r+1=4$ (ignoring the negative root because $r>0$). This gives $2r=3$, so $r=1.5$.

We have found the radius of the largest circle! To find the point of tangency $(\alpha ,\beta )$, we substitute $r=1.5$ back into our quadratic equation:

This is a perfect square: $(x-2{)}^2=0$. Thus, $x=2$, which means $\alpha =2$.

The elegance of this result—that the tangency occurs exactly at $x=2$—is a testament to the harmony of coordinate geometry. By transforming a geometric constraint into a simple quadratic discriminant problem, we have conquered the challenge.