AP EAMCET202318 May 2023Morning ShiftMathematicsApplication of DerivativesActual
If the locus of the points on the curve x^3 y^2+ x^2 y =5 at which the tangent is parallel to X -axis is f ( x , y )=0 , then the point that lies on this curve f(x, y)=0 is
Options
- A(2, [3] 3 )
- B( [3] 2 , 3)
- C(-2, 1 [3] 3 )
- D(- [3] 2 , 1 [3] 3 )
Correct answer
C. (-2, 1 [3] 3 )
Step-by-step solution
x^3 y^2+ x^2 y =5 Differentiating both sides with respect to ' x ', we get : 3 x^2 y^2+2 x^3 y d y d x + 2 x y - x^2 y^2 d y d x =0 aligned & (2 x^3 y- x^2 y^2 ) d y d x =- (3 x^2 y^2+ 2 x y ) & d y d x = y x ( 3 x y^3+2 2 x y^3-1 ) aligned Tangent is parallel to x -axis. aligned & d y d x =0 y x ( 3 x y^3+2 2 x y^3-1 )=0 & 3 x y^3+2=0 aligned The point satisfying above equation, will lies on f(x, y)=0 Now, let us check point (2, [3] 3 ): 6 3+2 0 Point ( [3] 2 , 3): 3 [3] 2 27+2 0 Point (-2, 1 [3] 3 ): 3(-2) 1 3 +2