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Summation by Sigma Method

If x=2t1+t2...

Question

If $x = \frac{2 t}{1 + t^{2}}$ and $y = \frac{1 - t^{2}}{1 + t^{2}}$ then eliminate $t$ .

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Solution

Given,
$x = \frac{2 t}{1 + t^{2}}$ ......(1) and $y = \frac{1 - t^{2}}{1 + t^{2}}$ ......(2).
Now, squaring (1) and (2) and then adding we get,
$x^{2} + y^{2} = \frac{4 t^{2}}{(1 + t^{2})^{2}} + \frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}}$
or, $x^{2} + y^{2} = \frac{(1 - t^{2})^{2} + 4 t^{2}}{(1 + t^{2})^{2}}$
or, $x^{2} + y^{2} = \frac{(1 + t^{2})^{2}}{(1 + t^{2})^{2}}$
or, $x^{2} + y^{2} = 1$ .
This is the required expressing without $t$ .

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Similar questions

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 t}{1 + t^{2}}

, then prove that

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

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

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

\frac{d y}{d x}

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

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

t \in (- 1, 0)

, then the value of

4 \frac{d^{2} y}{d x^{2}} - \frac{3}{2} \frac{d y}{d x} - \frac{1}{2} \frac{y}{x}

(x, y) = (1, - 1)

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

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

then the value of

\frac{d^{2} y}{d x^{2}}

t = 0

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