Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is 2y x². If the curve passes through the centre of the circle x² + y² – 2x – 2y = 0, then its equation is:

(B) x² log e |y| = –2(x – 1)

(D) x log e |y| = –2(x – 1)

Limit, Continuity and Differentiability: JEE Main 2019 Mathematics question

JEE Main 2019 · 8 Apr, Evening shift · Q2771

JEE Main 20198 Apr, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is

2y \over x^2

. If the curve passes through the centre of the circle x² + y² – 2x – 2y = 0, then its equation is :

JEE Main 2019 · 8 Apr, Evening shift · Q2771

JEE Main 20198 Apr, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is

2y \over x^2

. If the curve passes through the centre of the circle x² + y² – 2x – 2y = 0, then its equation is :