If x-1-t21-t2-y-2t1-t2- then prove that dydx-xy-

Put #160; t=tanθ x=1 t^21+t^2=1 tan^2θ1+tan^2θ=cos2θ dxdθ= 2sin2θ y=2t1 t^2=2tanθ1+tan^2θ=sin 2θ dydθ=2cos2θ dydx=cosθ sinθ= xy ...

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Integration Using Substitution

If x=1-t21+...

Question

If $x = \frac{1 - t^{2}}{1 + t^{2}}, y = \frac{2 t}{1 + t^{2}}$ , then prove that $\frac{d y}{d x} = - \frac{x}{y}$ .

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Solution

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Similar questions

x = \frac{1 - t^{2}}{1 + t^{2}}

y = \frac{2 t}{1 + t^{2}}

\frac{d y}{d x}

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

\frac{d y}{d x}

t = 2

x = \frac{1 - t^{2}}{1 + t^{2}}

y = \frac{2 a t}{1 + t^{2}}

\frac{d y}{d x}

x

\frac{t^{2}}{1 + t^{2}}

y

\frac{2 a t}{1 + t^{2}}

\frac{d y}{d x}

x^{2} + y^{2} = t + \frac{1}{t}

x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}

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

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