Locus of the point given by the equations x -2 at -1- t 2- y - a 1- t 2-1- t 2-1 - t - 1 is aA- Straight lineB- CircleC- HyperbolaD- Ellipse

The correct option is B Circle Here x = 2at/1 + t^2 and y = a(1 t^2)/1 + t^2 Squaring and adding both, we get x^2 + y^2 = a^2. ...

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Parametric Form of Tangent: Hyperbola

Locus of the ...

Question

Locus of the point given by the equations $x = \frac{2 a t}{1 + t^{2}}, y = \frac{a (1 - t^{2})}{1 - t^{2}} (- 1 \leq t \leq 1)$ is a

Straight line

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Circle

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Ellipse

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Hyperbola

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Solution

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Similar questions

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

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

t

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

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

t

t

x = a \frac{1 - t^{2}}{1 + t^{2}}, y = b \frac{2 t}{1 + t^{2}}

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

y = t - \frac{1}{t}

x = \frac{a (1 - t^{2})}{1 + t^{2}}, y = \frac{2 a t}{1 + t^{2}}

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