AP EAMCET202318 May 2023Morning ShiftMathematicsApplication of DerivativesActual
If the tangent drawn to the curve (x^2+1 )(y-3)=x at a point P , lying in the first quadrant, is a horizontal line, then the equation of the normal at the point P is
Options
- Ax= 7 2
- Bx=1
- Cy= 7 2
- Dy=1
Correct answer
B. x=1
Step-by-step solution
(x^2+1 )(y-3)=x ... (i) aligned & Then, (x^2+1 ) d y d x +(2 x+0)(y-3)=1 & d y d x = 6 x-2 x y+1 x^2+1 aligned Tangent is a horizontal line. d y d x =0 6 x-2 x y+1=0 ...(ii) Solving equations (i) and (ii), we get x=1, y= 7 2 Slope of normal, m=- 1 d y d x =- 1 0 Equation of normal at (1, 7 2 ) is (y- 7 2 )= (- 1 0 )(x-1) (x-1)=0 x=1