Use implicit differentiation to find an equation of the tangent line to the curve at the given point.2x2 + xy + 2y2 = 5,  (1, 1)  (ellipse)

Sure, here are the steps to find the equation of the tangent line to the curve at the given point using implicit differentiation:

1. Differentiate both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, you need to use the chain rule.

   The derivative of 2x^2 is 4x.

   The derivative of xy is y + x(dy/dx) by the product rule.

   The derivative of 2y^2 is 4y(dy/dx) by the chain rule.

   The derivative of 5 is 0.

   So, the differentiated equation is: 4x + y + x(dy/dx) + 4y(dy/dx) = 0.

2. Solve the equation for dy/dx.

   Rearrange the equation to: dy/dx = -(4x + y) / (x + 4y).

3. Substitute the point (1, 1) into the equation to find the slope of the tangent line at that point.

   dy/dx = -(4(1) + 1) / (1 + 4(1)) = -5/5 = -1.

   So, the slope of the tangent line at the point (1, 1) is -1.

4. Use the point-slope form of the equation of a line to find the equation of the tangent line.

   The point-slope form is: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line.

   Substituting the point (1, 1) and the slope -1, we get: y - 1 = -1(x - 1).

5. Simplify the equation to get the equation of the tangent line.

   y - 1 = -x + 1, which simplifies to y = -x + 2.

So, the equation of the tangent line to the curve at the point (1, 1) is y = -x + 2.