A) \[a>0,\,\,b>0\]
B) \[a>0,\,\,b<0\]
C) \[a<0,\,\,b<0\]
D) None of these
Correct Answer: B
Solution :
We have, the given curve \[xy=1\] On differentiating w.r.t.\[x,\] we get \[x\frac{dy}{dx}+y=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-y}{x}=-\frac{1}{{{x}^{2}}}\] \[(\because \,\,xy=1)\] \[\Rightarrow -\frac{dy}{dx}={{x}^{2}}=\]the slope of normal But the given normal is \[ax+by+c=0\] It's slope is \[-\frac{a}{b}\] which must be equal to \[{{x}^{2}}\] as both are the slopes of normal \[\therefore \] \[{{x}^{2}}=-\frac{a}{b}\] \[\Rightarrow \] \[\frac{-a}{b}>0\] \[(\because \,{{x}^{2}}>0)\] \[\Rightarrow \] \[a>0,\,\,b<0\]or\[a<0,\,\,b>0\]
You need to login to perform this action.
You will be redirected in
3 sec
