Question

Which of the following equations represents a **parabola**?

• A. \( x = 4\cos t,\ y = 4\sin t \)
• B. \( x^2 - 2 = -2\cos t,\ y = \cos^2 \left( \frac{t}{2} \right) \)
• C. \( \sqrt{x} = \tan t,\ \sqrt{y} = \sec t \)
• D. \( x = \sqrt{1 - \sin t},\ y = \sin \left( \frac{t}{2} \right) + \cos \left( \frac{t}{2} \right) \)

Hint:

Parametric forms can define conics. Substitute to eliminate parameter and derive standard conic forms.

Answer 1

The Correct Option is B

We analyze each option. Option (1): \( x = 4\cos t,\ y = 4\sin t \) This is a circle: \( x^2 + y^2 = 16 \) Option (2): \[ x^2 - 2 = -2\cos t \Rightarrow \cos t = -\frac{x^2 - 2}{2} \] Also: \[ y = \cos^2 \left( \frac{t}{2} \right) = \frac{1 + \cos t}{2} \Rightarrow y = \frac{1 - \frac{x^2 - 2}{2}}{2} = \frac{1}{2} - \frac{x^2 - 2}{4} \Rightarrow y = -\frac{x^2}{4} + 1 \] Clearly a parabola. Option (3): \[ \sqrt{x} = \tan t,\ \sqrt{y} = \sec t \Rightarrow x = \tan^2 t,\ y = \sec^2 t \Rightarrow y = 1 + x \] So this is a straight line. Option (4): Not reducible to quadratic form easily and not a conic. % Final Answer: \[ \boxed{x^2 - 2 = -2\cos t,\ y = \cos^2 \left( \frac{t}{2} \right)} \Rightarrow y = -\frac{x^2}{4} + 1 \Rightarrow \text{Parabola} \]