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TS EAMCET · Maths · Circle

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

  1. A the length of the latusrectum of a parabola
  2. B the radius of a circle
  3. C the length of the transverse axis of a hyperbola
  4. D the length of the semi-major axis of an ellipse
Verified Solution

Answer & Solution

Correct Answer

(B) the radius of a circle

Step-by-step Solution

Detailed explanation

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.