The equation of the tangent to the curve x^2 + 3y^2 = 4 at the point (1, 1) is
Correct: A
The given curve is x^2 + 3y^2 = 4. Differentiating both sides with respect to x, we get 2x + 6y*(dy/dx) = 0. So, dy/dx = -x/3y. At the point (1, 1), the slope is -1/3. The equation of the tangent is y - 1 = (-1/3)*(x - 1), which simplifies to 3y - 3 = -x + 1, or x + 3y - 4 = 0. None of the options match this equation exactly. The closest is 2x + 6y - 8 = 0