A) \[20x+y=140\]
B) \[20x-y=7\]
C) \[x-20y=7\]
D) \[20x-y=140\]
Correct Answer: A
Solution :
Equation of given curve is \[y(x-2)(x-3)-x+7=0\] ?(i) This curve cuts the .y-axis, where y = 0 \[\therefore \]\[0(x-2)(x-3)-x+7=0\] \[\Rightarrow \] x = 7 The point is (7,0). On differentiating Eq. (i), we get \[\frac{dy}{dx}(x-2)(x-3)+y(x-3+x-2)-1=0\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{1-y(2x-5)}{(x-2)(x-3)}\] \[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{(7,0)}}=\frac{1-0(14-5)}{(7-2)(7-3)}=\frac{1}{20}\] \[\therefore \]The equation of the normal is \[y-0=-20(x-7)\]or\[y+20x=140\]You need to login to perform this action.
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