MHT CET20242 May 2024Morning ShiftMathematicsApplication of DerivativesActual
The normal to the curve, y(x-2)(x-3)=x+6 at the point, where the curve intersects the Y-axis, passes through the point
Options
- A(- 1 2 ,- 1 2 )
- B( 1 2 , 1 2 )
- C( 1 2 ,- 1 3 )
- D( 1 2 , 1 3 )
Correct answer
B. ( 1 2 , 1 2 )
Step-by-step solution
Given equation of curve aligned & y(x-2)(x-3)=x+6 & y= x+6 (x-2)(x-3) & y= x+6 x^2-5 x+6 aligned Differentiating w.r.t. x , we get aligned & d y ~d x = (x^2-5 x+6 )(1)-(x+6)(2 x-5) (x^2-5 x+6 )^2 & = x^2-5 x+6-(x+6)(2 x-5) (x^2-5 x+6 )^2 & ~d y ~d x = -x^2-12 x+36 (x^2-5 x+6 )^2 & At Y-axis, x=0 aligned aligned d y ~d x_ at x=0 & = -(0)^2-12(0)+36 (0^2-5(0)+6 )^2 & = 36 36 & =1 aligned The equation of normal is y-1=-1(x-0) i.e., x+y=1 Option (B) i.e., ( 1 2 , 1 2 ) satisfies above equation Normal passes through ( 1