Question

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:?

Answer 1

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)

Answer 2

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