A) \[x+5y=2\]
B) \[x-5y=2\]
C) \[5x-y=2\]
D) \[5x+y-2=0\]
E) \[x-5y=0\]
Correct Answer: A
Solution :
The given equation of curve is\[(1+{{x}^{2}})y=2-x,\]it meets\[x-\]axis at (2, 0) Or it can be rewritten as\[y=\frac{2-x}{1+{{x}^{2}}}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{(1+{{x}^{2}})(-1)-(2-x)(2x)}{{{(1+{{x}^{2}})}^{2}}}\] \[=\frac{-1-{{x}^{2}}-4x+2{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}\] \[{{\left( \frac{dy}{dx} \right)}_{(2,0)}}=\frac{-1-4-8+8}{25}=-\frac{1}{5}\] \[\therefore \]Required equation of tangent is \[(y-0)=-\frac{1}{5}(x-2)\] \[\Rightarrow \] \[x+5y=2\]
You need to login to perform this action.
You will be redirected in
3 sec
