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The tangent to the hyperbola $xy = c^2$ at the point $P$ intersects the $x-$ axis at $T$ and the $y-$ axis at $T'$. The normal to the hyperbola at $P$ intersects the $ x-$ axis at $N$ and the $y-$ axis at $N'$. The areas of the triangles $PNT$ and $PN'T' $ are $ \Delta$ and $ \Delta ' $ respectively, then $\frac{1}{\Delta }\,\, + \,\,\frac{1}{{\Delta '}}\,$ is
equal to $ 1$
depends on $ t$
depends on $c$
equal to $2$
Solution
Tangent : $\frac{x}{{ct}} + \frac{{yt}}{c} = 2$
put $y = 0$ $x = 2ct (T)$
$x = 0$ $y = \frac{{2c}}{t}(T')$
paralleily normal is $y -\frac{c}{t}$ $= t^2(x – ct)$
put $y = 0$ $x = ct – \frac{c}{{{t^3}}}(N)$
$x = 0 $ $\frac{c}{t}$ $ – ct^3 (N')$
Area of $\Delta PNT$ $=\frac{c}{{2t}}\left( {ct + \frac{c}{{{t^3}}}} \right)$ $\Rightarrow$ $\Delta$ $=$ $\frac{{{c^2}(1 + {t^4})}}{{2{t^4}}}$
Area of $ \Delta PN'T'$ $=$ $ct\left( {\frac{c}{t} + c{t^3}} \right)$ $\Rightarrow$ $\Delta ' $ $ = $ $\frac{{{c^2}(1 + {t^4})}}{2}$
$\therefore$ $\frac{1}{\Delta } + \frac{1}{{\Delta '}}$ $=$ $\frac{{2{t^4}}}{{{c^2}(1 + {t^4})}} + \frac{2}{{{c^2}(1 + {t^4})}}$ $= $ $\frac{2}{{{c^2}(1 + {t^4})}}$ $(t^4 + 1) $ $=$ $\frac{2}{{{c^2}}}$
which is independent of $t.$
