Question

The points of intersection of the curves whose parametric equations are x = t² + 1, y = 2t and x = 2s, y = 2/s is given by

• A. (1, -3)
• B. (2, 2)
• C. (-2, 4)
• D. (1, 2)

Answer 1

Answer: (B)

(2, 2)

Answer 2

Eliminating t from x = t² + 1, y = 2t, we obtain y² = 4x - 4. Substituting x = 2s, y = $\frac{2}{s}$ in y² = 4x - 4, we obtain

2s³ - s² - 1 = 0 ⇒ (s - 1) (2s² + s + 1) = 0 ⇒ s = 1

Putting s = 1 in x = 2s, y = 2/s, we obtain x = 2, y = 2.