Question

Let $ABCD$ be a square of side length $2$ units. $C_2$ is the circle through vertices $A,B,C,D$ and $C_1$ is the circle touching all the sides of the square $ABCD.$ $L$ is a line through $A.$

A line $M$ through $A$ is drawn parallel to $BD$. Point $S$ moves such that its distances from the line $BD$ and the vertex $A$ are equal. If locus of $S$ cuts $M$ at $T_2$ and $T_3$ and $AC$ at $T_1$, then area of $\Delta T_1{T}_2{T}_3$ is

• A. $\frac{1}{2}$ sq. unit
• B. $\frac{2}{3}$ sq. units
• C. $1$ sq. units
• D. $2$ sq. units

Answer 1

The correct option is C $1$ sq. units

$AG=\sqrt{2}$
${AT}_1=T_{1}G=\frac{1}{\sqrt{2}}$

[as A is the focus, $T_1$ is the vertex and BD is the directrix of parabola].

Also $T_2{T}_3$ is latus rectum.

$T_2{T}_3=4\times \frac{1}{\sqrt{2}}$

Area of $\Delta T_1{T}_2{T}_3=\frac{1}{2}\times \frac{1}{\sqrt{2}}\times \frac{4}{\sqrt{2}}=1$

Hence. option C.