JEE Advanced · Mathematics · 13. Parabola

Paragraph:
Read the following passage and answer the questions.
Let \(A B C D\) 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 square \(A B C D\). \(L\) is the line through \(A\).Question:
A line \(M\) through \(A\) is drawn parallel to \(B D\). Points \(S\) moves such that its distances from the line \(B D\) are 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 unit

Verified Solution

Answer & Solution

Correct Answer

(C)
1 sq unit

Step-by-step Solution

Detailed explanation

\(\because \quad A G=\sqrt{2}\)
\[
\therefore \quad A T_1=T_1 G=\frac{1}{\sqrt{2}}
\]
\{ as \(A\) is the focus, \(T_1\) is the vertex and \(B D\) is the directrix of parabola.\}
Also, \(T_2 T_3\) is latus rectum
\[
\begin{aligned}
& \therefore \quad T_2 T_3=4 \cdot \frac{1}{\sqrt{2}} \\
& \therefore \text { Area of } \Delta T_1 T_2 T_3=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=1 \text { sq unit }
\end{aligned}
\]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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	List I		List II
A.	a =	P.	13
B.	b =	Q.	-3
C.	c =	R.	1
D.	d =	S.	-2