Use implicit differentiation to find an equation of the tangent line to the curve at the given point
x^2+2xy+4y^2=12. Point (2,1). (Eclipse)
x^2+2xy+4y^2=12
2x + 2y + 2xy' + 8yy' = 0
(x+4y)y' = -x-y
y' = -(x+y)/(x+4y)
at (2,1) y' = -1/2
so, the line is
y-1 = -1/2 (x-2)
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To find the equation of the tangent line to the curve at the given point, we can use implicit differentiation. Here's how you can do it step by step:
1. Start with the given equation: x^2 + 2xy + 4y^2 = 12.
2. Differentiate both sides of the equation with respect to x. Remember that when differentiating y with respect to x, we have to use the chain rule. The derivative of y with respect to x is dy/dx.
3. Differentiate the left-hand side of the equation using the product rule and chain rule. The derivative of x^2, 2xy, and 4y^2 with respect to x will be: 2x + 2y(dy/dx) + 8y(dy/dx).
4. Set the derivative equal to 0 to find the slope of the tangent line. In other words, find the value of dy/dx that satisfies the equation when x = 2 and y = 1.
2x + 2y(dy/dx) + 8y(dy/dx) = 0
5. Simplify the equation by combining like terms:
2x + 2y(dy/dx) + 8y(dy/dx) = 0
2x + 10y(dy/dx) = 0
6. Substitute the values x = 2 and y = 1 into the equation:
2(2) + 10(1)(dy/dx) = 0
4 + 10(dy/dx) = 0
10(dy/dx) = -4
dy/dx = -4/10 = -2/5
So, the slope of the tangent line at the point (2, 1) is -2/5.
7. Use the point-slope form of a line to find the equation of the tangent line. The point-slope form is: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Using the point (2, 1) and slope -2/5, the equation of the tangent line is:
y - 1 = (-2/5)(x - 2)
Now, you can simplify this equation further if needed.