Applications-Of-Derivatives Question 444
Question: The slope of tangent to the curve $ x=t^{2}+3t-8 $ , $ y=2t^{2}-2t-5 $ at the point (2, ?1) is
[MNR 1994]
Options:
A) $ \frac{22}{7} $
B) $ \frac{6}{7} $
C) ? 6
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ t=2 $ for the point (2,?1) $ \frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac{6}{7} $ for $ t=2 $ .
BETA
