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Question

If the normal at $$\left(ct,\dfrac{c}{t}\right)$$ on the curve $$xy=c^{2}$$ meets the curve again in $$t'$$, then 


A
t=1t3
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B
t=1t
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C
t=1t2
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D
t2=1t2
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Solution

The correct option is A $$t'=-\dfrac{1}{t^{3}}$$
$$ xy = c^{2}$$ and point (ct, c/t)
Let us equation of the normal
$$ y = \frac{c^{2}}{x}$$
$$ \frac{dy}{dx} = \frac{-c^{2}}{x^{2}}$$
hence slope of normal is given by $$ \frac{x^{2}}{c^{2}}$$ & (ct , c/t) is
$$ t^{2}$$ and its equation is
$$ y = t^{2}(x-ct) + c/t$$
$$ \therefore xt^{3}-yt-ct^{4}+c = 0$$
At this passes through $$ (ct^{1},c/t^{1})$$
$$ ct^{1}.t^{3} - \frac{c}{t^{1}}.t-ct^{4}+c = 0 $$
$$ ct^{3}.(t^{1}-t)+\frac{c}{t^{1}}(t^{1}-t) = 0$$
$$ (ct^{3}+\frac{c}{t^{1}}) (t^{1}-t) = 0$$
$$ ct^{3}+ \frac{c}{t^{1}} = 0 $$ $$ \rightarrow ct^{3} = \frac{c}{t^{1} }$$
$$ t^{3}= \frac{-1}{t^{1}}$$
$$ \rightarrow t^{1} = \frac{-1}{t^{3}}$$

1167871_1248180_ans_0c334639e53040f288bed9e962b3b39b.jpg

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