Conic Sections Question 10
Question: The points of intersection of the curves whose parametric equations are $ x=t^{2}+1,\ y=2t $ and $ x=2s,\ y=\frac{2}{s} $ is given by
Options:
A) $ (1,\ -3) $
B) (2, 2)
C) (-2, 4)
D) (1, 2)
Show Answer
Answer:
Correct Answer: B
Solution:
Eliminating t from $ x=t^{2}+1,y=2t, $ we obtain $ y^{2}=4x-4 $
Similarly eliminating s from $ x=2s,y=\frac{2}{s}, $ we get $ xy=4. $
Hence point of intersection is (2, 2).
BETA
