Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Imagine you are standing in the center of a perfectly symmetrical square, $ABCD$, with its center at the origin $(0,0)$. By placing the vertices at $A(1,1)$, $B(-1,1)$, $C(-1,-1)$, and $D(1,-1)$, we unlock a hidden simplicity.

The sum of squared distances from any point $P(x,y)$ to these four vertices is defined as $S=P{A}^2+P{B}^2+P{C}^2+P{D}^2$. Expanding this expression, the linear terms vanish, leaving us with the elegant result:

For any point on the incircle $C_1$, we have $x^2+y^2=1$, which yields a sum of $12$. For the circumcircle $C_2$, we have $x^2+y^2=2$, which yields a sum of $16$. The ratio is simply:

Nature's Equidistant Curve

Now, let us consider a circle with center $C^{\prime }$ and radius $r$ that touches the incircle $C_1$ externally and a fixed line $L$. The distance from $C^{\prime }$ to the origin is $r+1$, and its distance to $L$ is $r$.

If we shift $L$ by one unit to create a new line $L^{\prime }$, the distance from $C^{\prime }$ to $L^{\prime }$ becomes $r+1$. Consequently, $C^{\prime }$ is equidistant from a fixed point (the origin) and a fixed line ($L^{\prime }$).

By definition, this locus is a parabola. This is a classic JEE trap: always look for the equidistant property to identify conics.

Area of the Triangle

Finally, we consider a point $S$ moving such that its distance from line $BD$ equals its distance from vertex $A(1,1)$. This is another parabola where the focus is $A(1,1)$ and the directrix is the line $BD:x+y=0$.

The vertex $T_1$ is the midpoint of $A$ and the origin, which is $(\frac{1}{2},\frac{1}{2})$. The line $L^{\prime }$ passing through $A$ parallel to $BD$ acts as the latus rectum.

The distance from $A$ to $BD$ is $2a=\sqrt{2}$, which implies $a=\frac{1}{\sqrt{2}}$. The base of our triangle $T_1{T}_2{T}_3$ is the latus rectum, $4a=2\sqrt{2}$, and the height is $a=\frac{1}{\sqrt{2}}$.

The area is calculated as:

The final area is 1 square unit. You have successfully navigated the symmetry, identified the conic, and calculated the area with precision.