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:?
Calculus. I need help! - drwls, Friday, March 2, 2012 at 11:18pm
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)
Calculus. I need help! - MILY, Sunday, March 4, 2012 at 4:39pm
oops my bad.. 26(x^2-y^2)