Question

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

• A. $ t=-\frac{1}{{{t}^{3}}} $
• B. $ t=-\frac{1}{t} $
• C. $ t=\frac{1}{{{t}^{2}}} $
• D. $ t{{}^{2}}=-\frac{1}{{{t}^{2}}} $

Answer 1

Answer: (A)

The equation of normal at $ \left( ct,\frac{c}{t} \right) $ on the curve $ xy={{c}^{2}} $ is $ ty={{t}^{3}}x-c{{t}^{4}}+c $ If it passes through $ \left( ct,\frac{c}{t} \right) $ then $ t\frac{c}{t}={{t}^{3}}.ct-c{{t}^{4}}+c $ $ \Rightarrow $ $ t={{t}^{3}}t{{}^{2}}-{{t}^{4}}t+t $ $ \Rightarrow $ $ t-t={{t}^{3}}t(t-t) $ $ \Rightarrow $ $ 1=-{{t}^{3}}t $ $ \Rightarrow $ $ t=-\frac{1}{{{t}^{3}}} $