A line $M$ through $A$ is drawn parallel to $B D$. Point $S$ moves such that its distances from the

Question

A line $M$ through $A$ is drawn parallel to $B D$. Point $S$ moves such that its distances from the line $B D$ and the vertex $A$ are equal. If locus of $S$ cuts $M$ at $T_{2}$ and $T_{3}$ and $A C$ at $T_{1}$, then area of $\Delta T_{1} T_{2} T_{3}$ is\

Professor PC Thomas Classes · Answer

(c) : Let $A$ be the vertex and $T_{1}$ be the focus of the parabola\ As $A G=\sqrt{2}, \quad A T_{1}=T_{1} N=\frac{1}{\sqrt{2}}=a$ (say)\ Also $T_{2} T_{3}=$ Length of latus rectum\ $=4 a=4 	imes \frac{1}{\sqrt{2}}$\ $	herefore \quad$ Area of $\Delta T_{1} T_{2} T_{3}=\frac{1}{2} 	imes \frac{1}{\sqrt{2}} 	imes \frac{4}{\sqrt{2}}$\ $=1$ sq. unit.\