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A point on the curve $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ is
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Verified Answer
The correct answer is:
$(A \sec \theta, B \tan \theta)$
\(\begin{aligned}
& \text {If }(x, y) \equiv(A \sec \theta, B \tan \theta) \\
& \text {So, } \frac{(A \sec \theta)^2}{A_2}-\frac{(B \tan \theta)^2}{B^2}=1 \\
& \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1 \text {(identity) }
\end{aligned}\)
& \text {If }(x, y) \equiv(A \sec \theta, B \tan \theta) \\
& \text {So, } \frac{(A \sec \theta)^2}{A_2}-\frac{(B \tan \theta)^2}{B^2}=1 \\
& \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1 \text {(identity) }
\end{aligned}\)
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