2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    The equation of the normal to the curve \[y(x-2)(x-3)-x+7=0\]at the point where it cuts the x-axis is

    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.
You will be redirected in 3 sec spinner