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
B
t=1t
C
t=1t2
D
t2=1t2

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 $$(ct^{1},c/t^{1})$$$$ct^{1}.t^{3} - \frac{c}{t^{1}}.t-ct^{4}+c = 0$$$$ct^{3}.(t^{1}-t)+\frac{c}{t^{1}}(t^{1}-t) = 0$$$$(ct^{3}+\frac{c}{t^{1}}) (t^{1}-t) = 0$$$$ct^{3}+ \frac{c}{t^{1}} = 0$$ $$\rightarrow ct^{3} = \frac{c}{t^{1} }$$$$t^{3}= \frac{-1}{t^{1}}$$$$\rightarrow t^{1} = \frac{-1}{t^{3}}$$Maths

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