A curve C given by x = g(y) passes through the point (1, 1) . The tangent to C at any point P(x, y) intersects the x -axis at a point whose x -coordinate is y^2 x^2 . Then the value of g(3) is equal to:

Step-by-step solution

The equation of the tangent to the curve at point (x, y) is given by: Y - y = dy dx (X - x) To find the x -intercept, we set Y = 0 : -y = dy dx (X - x) X = x - y dx dy It is given that the x -intercept is y^2 x^2 . Thus, x - y dx dy = y^2 x^2 y dx dy - x = -y^2 x^2 Dividing the entire equation by x^2 y , we obtain a Bernoulli differential equation in x : 1 x^2 dx dy - 1 xy = -y Let u = 1 x , which implies du dy = - 1 x^2 dx dy . Substituting this into the equation gives: - du dy - u y = -y du dy + 1 y u = y This is