JEE Main 202310 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The slope of tangent at any point (x, y) on a curve y=y(x)y=y(x)y=y(x) is x2+y22xy,x>0{{{x^2} + {y^2}} \over {2xy}},x > 02xyx2+y2,x>0. If y(2)=0y(2) = 0y(2)=0, then a value of y(8)y(8)y(8) is :AOption A. −42- 4\sqrt 2−42BOption B. 232\sqrt 323COption C. 434\sqrt 343DOption D. −23- 2\sqrt 3−23CheckSkip · show solution
