If x - a t - 1t - y - a t - 1t Show that- dydx - xy-

x=a(t 1t); y=a(t,1t) x=a(t^2 1t) ;y=a(t^2+1t) dxdt=a(t(2t) (t^2 1)t^2) ;dydt=a(t(2t) (t^2+1)t^2) =a(t^1+1t^2); #160; dydt=a(t^2 1t^2) ...

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Parametric Differentiation

If x = a t ...

Question

If $x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})$
Show that: $\frac{d y}{d x} = \frac{x}{y}$ .

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Solution

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Similar questions

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

\frac{d y}{d x} = \frac{x}{y}

x = \sqrt{a^{{sin}^{- 1} t}}, y = \sqrt{a^{{cos}^{- 1} t}}

\frac{d y}{d x} = - \frac{y}{x}

x = \sqrt{a^{s i n^{- 1} t}}

y = \sqrt{a^{c o s^{- 1} t}}

\frac{d y}{d x} = \frac{- y}{x}

x = \frac{a}{1 + t^{3}}, y = \frac{a t}{1 + t^{3}}

\frac{d y}{d x}

t = 1

\frac{1}{p}

p =

x = \sqrt{a^{{sin}^{- 1} t}}, y = \sqrt{a^{{cos}^{- 1} t}}

\frac{d y}{d x} = - \frac{y}{x} .

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