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If $x=\frac{2 a t}{1+t^2}, y=\frac{a\left(1-t^2\right)}{1+t^2}$, where $t$ is a parameter, then $a$ is
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Verified Answer
The correct answer is:
the radius of a circle
We have,
$$
\begin{aligned}
& x=\frac{2 a t}{1+t^2} \\
& y=\frac{a\left(1-t^2\right)}{1+t^2}
\end{aligned}
$$
put $t=\tan \theta$, We get
$$
x=a \sin 2 \theta, y=a \cos 2 \theta
$$
Squaring and adding we get
$$
x^2+y^2=a^2
$$
$\therefore a$ is radius of circle.
$$
\begin{aligned}
& x=\frac{2 a t}{1+t^2} \\
& y=\frac{a\left(1-t^2\right)}{1+t^2}
\end{aligned}
$$
put $t=\tan \theta$, We get
$$
x=a \sin 2 \theta, y=a \cos 2 \theta
$$
Squaring and adding we get
$$
x^2+y^2=a^2
$$
$\therefore a$ is radius of circle.
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