Question

The locus of the point of intersection of the lines \(x = a(1 - t^2)/(1 + t^2)\) and \(y = 2at/(1 + t^2)\) (t being a parameter) represents:

• A. Circle
• B. Parabola
• C. Ellipse
• D. Hyperbola

Hint:

Understanding the geometrical interpretation of parametric equations simplifies the identification of loci.

Answer 1

The Correct Option is A

Given that: \[ x = a \left( \frac{1 - t^2}{1 + t^2} \right) \quad {and} \quad y = \frac{2at}{1 + t^2} \] Let \( t = \tan \theta \). Thus: \[ x = a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \quad {and} \quad y = \frac{2a \tan \theta}{1 + \tan^2 \theta} \] From this, we have: \[ x = a \cos 2\theta \quad {and} \quad y = a \sin 2\theta \] We can write: \[ \cos 2\theta = \frac{x}{a} \quad {and} \quad \sin 2\theta = \frac{y}{a} \] Squaring both sides: \[ \cos^2 2\theta = \frac{x^2}{a^2} \quad {and} \quad \sin^2 2\theta = \frac{y^2}{a^2} \] Adding these equations: \[ \cos^2 2\theta + \sin^2 2\theta = \frac{x^2}{a^2} + \frac{y^2}{a^2} \] Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we get: \[ 1 = \frac{x^2 + y^2}{a^2} \] Thus, the equation becomes: \[ x^2 + y^2 = a^2 \] This represents the equation of a circle with center at the origin and radius \( a \). Therefore, the locus of the point is a circle having center at the origin and radius \( a \).