HARDER PARTS WAS 3(x^2+y^2)^2=26(y^2+y^2)

Find the equation of the tangent line to the curve (a lemniscate) 3(x^2+y^2)^2=26(y^2+y^2) at the point (-4,2). The equation of this tangent line can be written in the form y=mx+b where m is:? and where b is:?

The slope m is the value of dy/dx at x = -4 and y = 2.

You can get dy/dx by implicit differentiation.

Arre you sure your last term is (y^2 + y^2) ? Why wouldn't they write that as 2 y^2? The equation you wrote is not satisfied at (-4,2)

oops my bad.. 26(x^2-y^2)

To find the equation of the tangent line to a curve, we need to first find the derivative of the curve. Then, we can substitute the coordinates of the given point into the derivative to find the slope of the tangent line. Finally, we can plug the slope and the coordinates of the given point into the equation of a line (y=mx+b) to find the value of b.

Let's follow these steps:

Step 1: Find the derivative of the curve
Differentiate both sides of the equation with respect to x. Let's simplify the equation first:

3(x^2+y^2)^2 = 26(y^2+y^2)
3(x^4 + 2x^2y^2 + y^4) = 26(2y^2)
3x^4 + 6x^2y^2 + 3y^4 = 52y^2

Now, differentiate both sides with respect to x using the power rule for derivatives:

12x^3 + 12xy^2 + 6y^2x^2(dy/dx) + 12y^3(dy/dx) = 0

Step 2: Find the slope of the tangent line
Substitute the coordinates of the given point (-4,2) into the derivative equation:

12(-4)^3 + 12(-4)(2)^2 + 6(2)^2(-4)^2(dy/dx) + 12(2)^3(dy/dx) = 0

Simplify this equation to solve for dy/dx, which represents the slope of the tangent line at (-4,2).

Step 3: Plug the values into the equation of the line
Now that we have the slope (m), we can plug it and the coordinates of the given point (-4,2) into the equation of a line (y=mx+b) to find the value of b.

Thus, the equation of the tangent line to the curve 3(x^2+y^2)^2=26(y^2+y^2) at the point (-4,2) can be written as y = mx + b, where m is the slope of the tangent line and b is the y-intercept.