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If the normal to the rectangular hyperbola $x y=$ $\mathrm{c}^{2}$ at the point $\left(\mathrm{ct}, \frac{\mathrm{c}}{\mathrm{t}}\right)$ meets the curve again at $\left(\mathrm{ct}^{\prime}, \frac{\mathrm{c}}{\mathrm{t}^{\prime}}\right)$, then
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Verified Answer
The correct answer is:
$\mathrm{t}^{3} \mathrm{t}^{\prime}=-1$
The equation of tangent at $\left(\mathrm{ct}, \frac{\mathrm{c}}{\mathrm{t}}\right)$ is
$\mathrm{ty}=\mathrm{t}^{3} \mathrm{x}-\mathrm{ct}^{4}+\mathrm{c}$
If it passes through $\left(\mathrm{ct}^{\prime}, \frac{\mathrm{c}}{\mathrm{t}^{\prime}}\right)$ then
$\begin{array}{l}
\Rightarrow \frac{t c}{t^{\prime}}=t^{3} c t^{\prime}-c t^{4}+c \\
\Rightarrow t=t^{3} t^{\prime 2}-t^{4} t^{\prime}+t^{\prime} \\
\Rightarrow t \cdot t^{\prime}=t^{3} t^{\prime}\left(t^{\prime} . t\right) \Rightarrow t^{3} t^{\prime}=-1
\end{array}$
Note : If we take the co-ordinate axes along the asymptotes of a rectangular hyperbola, then the general equation $x^{2}-y^{2}=a^{2}$ becomes $x y=c^{2}$, where $c$ is a constant.
$\mathrm{ty}=\mathrm{t}^{3} \mathrm{x}-\mathrm{ct}^{4}+\mathrm{c}$
If it passes through $\left(\mathrm{ct}^{\prime}, \frac{\mathrm{c}}{\mathrm{t}^{\prime}}\right)$ then
$\begin{array}{l}
\Rightarrow \frac{t c}{t^{\prime}}=t^{3} c t^{\prime}-c t^{4}+c \\
\Rightarrow t=t^{3} t^{\prime 2}-t^{4} t^{\prime}+t^{\prime} \\
\Rightarrow t \cdot t^{\prime}=t^{3} t^{\prime}\left(t^{\prime} . t\right) \Rightarrow t^{3} t^{\prime}=-1
\end{array}$
Note : If we take the co-ordinate axes along the asymptotes of a rectangular hyperbola, then the general equation $x^{2}-y^{2}=a^{2}$ becomes $x y=c^{2}$, where $c$ is a constant.
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