Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is [2y x^2].

Question

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 :

Accepted Answer

Slope, $${{dy} \over {dx}}$$ = $$2y \over x^2$$

$$ \Rightarrow $$ $$\int {{{dy} \over y}}  = \int {2{{dx} \over {{x^2}}}} $$

$$ \Rightarrow $$ $${\log _e}|y| =  - {2 \over x} + C$$ ....... (1)

Center of the circle x2 + y2 -- 2x -- 2y = 0 is (1, 1)

Equation (1) passes through point (1, 1)

$$ \therefore $$ 0 = -2 + C

$$ \Rightarrow $$ C = 2

$$ \therefore $$ $${\log _e}|y| =  - {2 \over x} + 2$$

$$ \Rightarrow $$ x$${\log _e}|y| =  - 2 + 2x$$

$$ \Rightarrow $$ x$${\log _e}|y| =  2(x - 1)$$

Solutions · 1