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 \( a \).

Answer 2

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

Solution:
We are given parametric equations:
\(x = a\frac{1 - t^2}{1 + t^2}\) and \(y = \frac{2at}{1 + t^2}\)

We aim to eliminate the parameter \(t\) to find the locus (relation between \(x\) and \(y\)).

Step 1: Let’s simplify the expressions:
Let’s denote:
\[ x = a\frac{1 - t^2}{1 + t^2}, \quad y = \frac{2at}{1 + t^2} \]

Step 2: Divide both equations by \(a\):
\[ \frac{x}{a} = \frac{1 - t^2}{1 + t^2}, \quad \frac{y}{a} = \frac{2t}{1 + t^2} \]

Let us denote \(X = \frac{x}{a}, Y = \frac{y}{a}\), then:
\[ X = \frac{1 - t^2}{1 + t^2}, \quad Y = \frac{2t}{1 + t^2} \]

Step 3: Use identity:
\[ X^2 + Y^2 = \left(\frac{1 - t^2}{1 + t^2}\right)^2 + \left(\frac{2t}{1 + t^2}\right)^2 \]
Compute numerator: \[ (1 - t^2)^2 + (2t)^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1 + t^2)^2 \] So: \[ X^2 + Y^2 = \frac{(1 + t^2)^2}{(1 + t^2)^2} = 1 \] Thus: \[ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 = 1 \Rightarrow \frac{x^2 + y^2}{a^2} = 1 \Rightarrow x^2 + y^2 = a^2 \] Hence, the locus is a circle of radius \(a\) centered at origin.

Correct Answer: Circle