Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Welcome, future engineers. Today, we are not just solving a problem; we are choreographing a dance between two of the most fundamental shapes in coordinate geometry: the circle and the parabola.

We have a circle, $x^2+y^2=9$, a perfect, symmetric loop centered at the origin with a radius of $3$. We also have a parabola, $y^2=8x$, a sweeping curve that opens to the right, also anchored at the origin.

Our goal is to find where they meet, how they behave, and the beautiful geometric structures that emerge from their interaction.

To find the points of intersection, $P$ and $Q$, we must solve the system of equations simultaneously:

The substitution is elegant: replace $y^2$ in the circle equation with $8x$. This transforms our problem into a simple quadratic equation:

Factoring this gives us $(x+9)(x-1)=0$. We have two roots, $x=1$ and $x=-9$.

However, for the parabola $y^2=8x$, $x$ must be non-negative for $y$ to be a real number. Therefore, $x=-9$ is physically impossible in this context and we discard it.

Substituting $x=1$ back into $y^2=8x$, we find $y^2=8$, which gives us $y=\pm 2\sqrt{2}$. Thus, our points of intersection are $P(1,2\sqrt{2})$ and $Q(1,-2\sqrt{2})$.

A tangent is a line that touches the curve at a single point. For the circle at $P(1,2\sqrt{2})$, we use the formula $x{x}_1+y{y}_1=r^2$:

This line cuts the x-axis at point $R$. To find $R$, we set $y=0$, yielding $x=9$. So, $R$ is at $(9,0)$.

Next, for the parabola, the tangent at $P(1,2\sqrt{2})$ follows the formula $y{y}_1=2a(x+x_1)$. With $4a=8$, we have $a=2$, and the equation becomes:

Setting $y=0$ to find the x-intercept $S$, we get $4(x+1)=0$, which means $x=-1$. Point $S$ is at $(-1,0)$.

We now have our vertices: $P(1,2\sqrt{2})$, $Q(1,-2\sqrt{2})$, $R(9,0)$, and $S(-1,0)$. We compare the areas of $\Delta PQS$ and $\Delta PQR$.

For $\Delta PQS$, the base $PQ$ is a vertical segment with length $|2\sqrt{2}-(-2\sqrt{2})|=4\sqrt{2}$. The height is the horizontal distance from the x-coordinate of $S$ (which is $-1$) to the x-coordinate of the base (which is $1$), giving a height of $2$.

For $\Delta PQR$, the base $PQ$ is $4\sqrt{2}$. The height is the horizontal distance from the x-coordinate of $R$ (which is $9$) to the x-coordinate of the base (which is $1$), giving a height of $8$.

The ratio of these areas is:

For the circumradius $R$ of $\Delta PRS$, we use the formula $R=\frac{abc}{4\Delta }$. The side lengths are $SR=10$, $PR=\sqrt{(9-1{)}^2+(0-2\sqrt{2}{)}^2}=\sqrt{72}=6\sqrt{2}$, and $PS=\sqrt{(1-(-1){)}^2+(2\sqrt{2}-0{)}^2}=\sqrt{12}=2\sqrt{3}$.

The area $\Delta$ of $\Delta PRS$ is $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 10\times 2\sqrt{2}=10\sqrt{2}$. Plugging these into the circumradius formula:

For the inradius $r$ of $\Delta PQR$, we use $r=\frac{\Delta }{s}$. The sides are $PQ=4\sqrt{2}$, $PR=6\sqrt{2}$, and $QR=6\sqrt{2}$. The semi-perimeter $s$ is $8\sqrt{2}$ and the area $\Delta$ is $16\sqrt{2}$.

We started with two abstract equations and ended with concrete geometric properties. This is the essence of JEE Advanced mathematics: visualizing the geometry, respecting the symmetry, and executing the algebra with precision.