Final answer:
By applying implicit differentiation to the ellipse equation and finding the slope dy/dx, we can construct the equation of the tangent line at the point (x0, y0) and confirm that it is indeed x0x/a^2 + y0y/b^2 = 1.
Step-by-step explanation:
To show that the tangent line to the ellipse x^2/a^2 + y^2/b^2 = 1 at the point (x0, y0) is given by x0x/a^2 + y0y/b^2 = 1, we use implicit differentiation. Differentiate both sides of the ellipse equation with respect to x, treating y as a function of x (y=y(x)).
1. The derivative of x^2/a^2 is 2x/a^2. 2. Applying the chain rule, the derivative of y^2/b^2 is 2y/b^2 dy/dx. 3. Differentiating the right side, the derivative of 1 is 0.
Thus, we obtain 2x/a^2 + 2y/b^2 (dy/dx) = 0. Solving for dy/dx, we find that dy/dx = - (a^2/b^2)(x/y). This is the slope of the tangent line at any point (x, y) on the ellipse.
To find the equation of the tangent line at (x0, y0), substitute x0 for x and y0 for y in the differentiation result, which gives us the slope of the tangent line at that point. Rearranging the terms and multiplying both sides by y/y0, we get y/y0 = - (a^2/b^2)(x/x0). Simplify and multiply both sides by (b^2/x0y0) to get x0x/a^2 + y0y/b^2 = 1, which confirms the tangent line equation.