Assertion: The distance between the points P( | vedclass

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(C) The given equation $9x^2 - 16y^2 = 9$ can be written as $\frac{x^2}{1^2} - \frac{y^2}{(3/4)^2} = 1$. Here $a=1, b=3/4$.The parametric points on a

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$(A)$ is true but $(R)$ is false

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(C) The given equation $9x^2 - 16y^2 = 9$ can be written as $\frac{x^2}{1^2} - \frac{y^2}{(3/4)^2} = 1$. Here $a=1, b=3/4$.The parametric points on a hyperbola are $(a \sec 	heta, b 	an 	heta)$.For $	heta = \frac{\pi}{4}$,$P_1 = (1 \cdot \sqrt{2}, \frac{3}{4} \cdot 1) = (\sqrt{2}, \frac{3}{4})$.For $	heta = \frac{\pi}{3}$,$P_2 = (1 \cdot 2, \frac{3}{4} \cdot \sqrt{3}) = (2, \frac{3\sqrt{3}}{4})$.The distance $D = \sqrt{(2 - \sqrt{2})^2 + (\frac{3\sqrt{3}}{4} - \frac{3}{4})^2} = \sqrt{(4 + 2 - 4\sqrt{2}) + \frac{9}{16}(3 + 1 - 2\sqrt{3})} = \sqrt{6 - 4\sqrt{2} + \frac{9}{4} - \frac{9\sqrt{3}}{8}} = \sqrt{\frac{48 - 32\sqrt{2} + 18 - 9\sqrt{3}}{8}} = \sqrt{\frac{66 - 32\sqrt{2} - 9\sqrt{3}}{8}} = \frac{1}{2\sqrt{2}} \sqrt{66 - 32\sqrt{2} - 9\sqrt{3}}$.The Assertion is true.The Reason states $x = a \cosh t, y = b \sinh t$ are parametric equations for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is incorrect; the correct parametric equations are $x = a \sec 	heta, y = b 	an 	heta$.Thus,$(A)$ is true but $(R)$ is false.

Assertion: The distance between the points $P(\frac{\pi}{4})$ and $P(\frac{\pi}{3})$ on the hyperbola $9x^2 - 16y^2 = 9$ is $\frac{1}{4} \sqrt{66 - 32\sqrt{2} - 18\sqrt{3}}$.
Reason: $x = a \cosh t, y = b \sinh t$ are the parametric equations of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

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Assertion: The distance between the points $P(\frac{\pi}{4})$ and $P(\frac{\pi}{3})$ on the hyperbola $9x^2 - 16y^2 = 9$ is $\frac{1}{4} \sqrt{66 - 32\sqrt{2} - 18\sqrt{3}}$.Reason: $x = a \cosh t, y = b \sinh t$ are the parametric equations of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.