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Assertion : The distance between the points $p\left(\frac{\pi}{4}\right)$ and $p\left(\frac{\pi}{3}\right)$ on the hyperbola $9 x^2+16 y^2=9$ is
$$
\frac{1}{2 \sqrt{2}} \sqrt{66-33 \sqrt{2}-9 \sqrt{3}}
$$
Reason : $x=a \cosh t, y=b \sin \mathrm{h} t$ are the parametric equations of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The correct option among the following is
Options:
$$
\frac{1}{2 \sqrt{2}} \sqrt{66-33 \sqrt{2}-9 \sqrt{3}}
$$
Reason : $x=a \cosh t, y=b \sin \mathrm{h} t$ are the parametric equations of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The correct option among the following is
Solution:
2101 Upvotes
Verified Answer
The correct answer is:
(A) is true but (R) is false
Given equation of hyper bola is $9 x^2+16 y^2=9$ which is not in the standard form of hyper bola. Equation of hyperbola should be $9 x^2-16 y^2=9$
$\frac{x^2}{(1)^2}-\frac{y^2}{\left(\frac{3}{4}\right)^2}=1$
So, $a=1$ and $b=\frac{3}{4}$.
$$
\text { Point } \begin{aligned}
\mathrm{P}\left(\frac{\pi}{4}\right) & =(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta) \\
& =\left(1 \sec \left(\frac{\pi}{4}\right), \frac{3}{4} \tan \frac{\pi}{4}\right) \\
& =\left(\sqrt{2}, \frac{3}{4}\right)
\end{aligned}
$$
Point $\mathrm{P}\left(\frac{\pi}{3}\right)=(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$
$$
\begin{aligned}
& =\left(1 \sec \left(\frac{\pi}{3}\right), \frac{3}{4} \tan \left(\frac{\pi}{3}\right)\right) \\
& =\left(2, \frac{3}{4} \sqrt{3}\right)
\end{aligned}
$$
$$
\begin{aligned}
& \text { Required distance } \mathrm{D}=\sqrt{(2-\sqrt{2})^2+\left(\frac{3}{4} \sqrt{3}-\frac{3}{4}\right)^2} \\
& \qquad \begin{aligned}
\mathrm{D} & =\sqrt{4+2-4 \sqrt{2}+\frac{27}{16}+\frac{9}{16}-\frac{18}{16} \sqrt{3}} \\
\mathrm{D} & =\sqrt{\frac{132-64 \sqrt{2}-18 \sqrt{3}}{16}}
\end{aligned}
\end{aligned}
$$
$$
\mathrm{D}=\frac{1}{2 \sqrt{2}} \sqrt{66-32 \sqrt{2}-9 \sqrt{3}}
$$
So, Asseration is true but Reason of the parametric equation of hyperbola is incorrect because $\mathrm{x}=\operatorname{asec} \theta$ and $\mathrm{y}=$ $\mathrm{b} \tan \theta$.
So, option (c) is correct
$\frac{x^2}{(1)^2}-\frac{y^2}{\left(\frac{3}{4}\right)^2}=1$
So, $a=1$ and $b=\frac{3}{4}$.
$$
\text { Point } \begin{aligned}
\mathrm{P}\left(\frac{\pi}{4}\right) & =(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta) \\
& =\left(1 \sec \left(\frac{\pi}{4}\right), \frac{3}{4} \tan \frac{\pi}{4}\right) \\
& =\left(\sqrt{2}, \frac{3}{4}\right)
\end{aligned}
$$
Point $\mathrm{P}\left(\frac{\pi}{3}\right)=(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$
$$
\begin{aligned}
& =\left(1 \sec \left(\frac{\pi}{3}\right), \frac{3}{4} \tan \left(\frac{\pi}{3}\right)\right) \\
& =\left(2, \frac{3}{4} \sqrt{3}\right)
\end{aligned}
$$
$$
\begin{aligned}
& \text { Required distance } \mathrm{D}=\sqrt{(2-\sqrt{2})^2+\left(\frac{3}{4} \sqrt{3}-\frac{3}{4}\right)^2} \\
& \qquad \begin{aligned}
\mathrm{D} & =\sqrt{4+2-4 \sqrt{2}+\frac{27}{16}+\frac{9}{16}-\frac{18}{16} \sqrt{3}} \\
\mathrm{D} & =\sqrt{\frac{132-64 \sqrt{2}-18 \sqrt{3}}{16}}
\end{aligned}
\end{aligned}
$$
$$
\mathrm{D}=\frac{1}{2 \sqrt{2}} \sqrt{66-32 \sqrt{2}-9 \sqrt{3}}
$$
So, Asseration is true but Reason of the parametric equation of hyperbola is incorrect because $\mathrm{x}=\operatorname{asec} \theta$ and $\mathrm{y}=$ $\mathrm{b} \tan \theta$.
So, option (c) is correct
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