The equation x=2aθ1+θ2,y=a1−θ21+θ2 where a is constant, is the parametric equation of the curve

:y=a(1−θ)21+θ2,⇒θ2=a−ya+yx2=4a2θ21+θ22=4a2a−ya+y1+a−ya+y2x2=(a−y)(a+y)=a2−y2or x2+y2=a2

The equation x=2aθ1+θ2,y=a1−θ21+θ2 where a is constant, is the parametric equation of the curve

:y=a(1−θ)21+θ2,⇒θ2=a−ya+yx2=4a2θ21+θ22=4a2a−ya+y1+a−ya+y2x2=(a−y)(a+y)=a2−y2or x2+y2=a2

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The equation $x = \frac{2 a θ}{1 + θ^{2}}, y = \frac{a (1 - θ^{2})}{1 + θ^{2}}$ where a is constant, is the parametric equation of the curve

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$x^{2} - y^{2} = a^{2}$

$x^{2} + y^{2} = a^{2}$

$x^{2} + 4 y^{2} = 4 a^{2}$

$x = 2 y$

answer is B.

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Detailed Solution

$\begin{aligned} : y = \frac{a (1 - θ)^{2}}{1 + θ^{2}}, \Rightarrow θ^{2} = \frac{a - y}{a + y} \\ x^{2} = \frac{4 a^{2} θ^{2}}{{(1 + θ^{2})}^{2}} = 4 a^{2} \frac{(\frac{a - y}{a + y})}{{(1 + \frac{a - y}{a + y})}^{2}} \end{aligned}$
$\begin{aligned} x^{2} = (a - y) (a + y) = a^{2} - y^{2} \\ o r x^{2} + y^{2} = a^{2} \end{aligned}$