Question:

The Cartesian form of the curve given by \( x = \frac{a}{2}\left(t + \frac{1}{t}\right) \), \( y = \frac{a}{2}\left(t - \frac{1}{t}\right) \), where \( t \) is a parameter, is:

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When converting parametric equations to Cartesian form, eliminate the parameter by manipulating and combining the equations for \( x \) and \( y \).
Updated On: May 13, 2025
  • \( x^2 + y^2 = a^2 \)
  • \( x^2 - y^2 = a^2 \)
  • \( 2x^2 - y^2 = a^2 \)
  • \( x^2 - 2y^2 = a^2 \)
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The Correct Option is B

Solution and Explanation

Step 1: Express \( x \) and \( y \) in terms of \( t \).
We are given: \[ x = \frac{a}{2}\left(t + \frac{1}{t}\right), \quad y = \frac{a}{2}\left(t - \frac{1}{t}\right) \]
Step 2: Find \( x^2 \) and \( y^2 \).
First, square both expressions for \( x \) and \( y \): \[ x^2 = \left( \frac{a}{2} \right)^2 \left(t + \frac{1}{t}\right)^2 = \frac{a^2}{4}\left(t^2 + 2 + \frac{1}{t^2}\right) \] \[ y^2 = \left( \frac{a}{2} \right)^2 \left(t - \frac{1}{t}\right)^2 = \frac{a^2}{4}\left(t^2 - 2 + \frac{1}{t^2}\right) \]
Step 3: Subtract \( y^2 \) from \( x^2 \).
Now subtract \( y^2 \) from \( x^2 \): \[ x^2 - y^2 = \frac{a^2}{4} \left( \left(t^2 + 2 + \frac{1}{t^2}\right) - \left(t^2 - 2 + \frac{1}{t^2}\right) \right) \] \[ x^2 - y^2 = \frac{a^2}{4} \times 4 = a^2 \] Hence, the correct equation is: \[ x^2 - y^2 = a^2 \]
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