Circle 2 Question 19
20. 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
(a) $\frac{1}{2}$ sq unit
(b) $\frac{2}{3}$ sq unit
(c) 1 sq unit
(d) 2 sq units
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Answer:
Correct Answer: 20. (c)
Solution:
- Since,
$ A G=\sqrt{2} $
$ \therefore \quad A T _1=T _1 G=\frac{1}{\sqrt{2}} $
As, $A$ is the focus, $T _1$ is the vertex and $B D$ is the directrix of parabola.
Also, $T _2 T _3$ is latusrectum.
$ \begin{array}{ll} \therefore T _2 T _3=4 \cdot \frac{1}{\sqrt{2}} \\ \therefore \quad \text { Area of } \Delta T _1 T _2 T _3=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=1 \text { sq unit } \end{array} $
