Question

The point 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

x = t² + 1, y = 2t ⇒ x – 1 = $\frac{y^{2}}{4}$

and x = 2s, y = 2/s ⇒ xy = 4.

For the point of intersection we have,

$\frac{4}{y} - 1 = \frac{y^{2}}{4}$ ⇒ y³ + 4y – 16 = 0 ⇒ y = 2 ⇒ x = 2