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Question: If the normal at \(\left( ct,\frac{c}{t} \right)\) on the curve \(xy = c^{2}\) meets the curve again...

If the normal at (ct,ct)\left( ct,\frac{c}{t} \right) on the curve xy=c2xy = c^{2} meets the curve again in t′, then

A

t′=−1t3t^{'} = - \frac{1}{t^{3}}

B

t′=−1tt^{'} = - \frac{1}{t}

C

t′=1t2t^{'} = \frac{1}{t^{2}}

D

t′2=−1t2t^{'2} = - \frac{1}{t^{2}}

Answer

t′=−1t3t^{'} = - \frac{1}{t^{3}}

Explanation

Solution

The equation of the tangent at (ct,ct)\left( ct,\frac{c}{t} \right) is ty=t3x−ct4+cty = t^{3}x - ct^{4} + c

If it passes through(ct′,ct′)\left( ct^{'},\frac{c}{t^{'}} \right) then

⇒ tct′=t3ct′−ct4+c\frac{tc}{t^{'}} = t^{3}ct^{'} - ct^{4} + c ⇒ t=t3t′2−t4t′+t′t = t^{3}t^{'2} - t^{4}t^{'} + t^{'}

⇒ t−t′=t3t′(t′−t)t - t^{'} = t^{3}t^{'}(t^{'} - t) ⇒ t′=−1t3t^{'} = - \frac{1}{t^{3}}