Let E be an ellipse with equation (x/3)^2 + (y/4)^2 = 1. Find the equation of the tangent to E at the point P where x = √3 and y > 0, in the form ax + by = c, where a, b and c are rational.

In order to find the equation of the tangent to a curve at a point (x₁, y₁), we use the equation y - y₁ = m(x - x₁), where m is the gradient of the curve at (x₁, y₁). We already have x₁ = √3, so we now must find y₁ and m.

Recall that an ellipse with equation (x/a)² + (y/b)² = 1 has parametric equations x = a cos(t) and y = b sin(t), so in this case, E is described by the parametric equations x = 3 cos(t) and y = 4 sin(t). Since x = √3, cos(t) = 1/√3, and hence sin(t) = √(2/3), which indicates that y₁ = 4√(2/3). We then use the identity dy/dx = (dy/dt) / (dx/dt). dy/dt = 4 cos(t) and dx/dt = -3 sin(t), hence dy/dx = -4/3 cot(t). Since sin(t) = √(2/3) and cos(t) = 1/√3, cot(t) = 1/√2, hence dy/dx at this point is -(2√2)/3, so m = -(2√2)/3.

Finally, we simply sub m, x₁ and y₁ into our equation and rearrange. y - 4√(2/3) = -(2√2)/3 * (x - √3), implying that 3y - 4√6 = 2√6 - (2√2)x, which we can rearrange into the required form: (2√2)x + 3y = 6√6, with a = 2√2, b = 3 and c = 6√6.