Applications-Of-Derivatives Question 469
Question: The equation of the tangent to the curve $ (1+x^{2})y=2-x, $ where it crosses the x-axis, is
[Kerala (Engg.) 2002]
Options:
A) $ x+5y=2 $
B) $ x-5y=2 $
C) $ 5x-y=2 $
D) $ 5x+y-2=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ (1+x^{2})y=2-x $ ……(i) It meets x-axis, where $ y=0 $ i.e., $ 0=2-x\Rightarrow x=2 $ So, (i) meets x-axis at the point (2, 0) Also from (i), $ y=\frac{2-x}{1+x^{2}} $
Þ $ \frac{dy}{dx}=\frac{(1+x^{2}),(-1)-(2-x),(2x)}{{{(1+x^{2})}^{2}}} $
Þ $ \frac{dy}{dx}=\frac{x^{2}-4x-1}{{{(1+x^{2})}^{2}}} $ Slope of tangent at (2, 0)is, $ \frac{4-8-1}{{{(1+4)}^{2}}}=\frac{-5}{25}=\frac{-1}{5} $ \ Equation of tangent at (2, 0) is , $ y-0=-\frac{1}{5}(x-2)\Rightarrow x+5y=2 $ .
BETA
