Applications Of Derivatives Question 49
Question: A curve is represented by the equation $ x=sec^{2}t $ and $ y=cott $ , where t is a parameter. If the tangent at the point P on the curve where $ t=\pi /4 $ meets the curve again at the point Q, then $ | PQ | $ is equal to
Options:
A) $ \frac{5\sqrt{3}}{2} $
B) $ \frac{5\sqrt{5}}{2} $
C) $ \frac{2\sqrt{5}}{3} $
D) $ \frac{3\sqrt{5}}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Eliminating t gives $ y^{2}(x-1)=1 $ . Equation of the tangent at $ P(2,1) $ is $ x+2y=4 $ .
Solving with curve $ x=5 $ and $ y=-1/2 $ , we get $ Q\equiv (5,-1/2) $ or $ PQ=\frac{3\sqrt{5}}{2} $
BETA
