Find the The Slopes of the tangent and the normal to the following curves at the indicated points :

x2 + 3y + y2 = 5 at (1, 1)

Given:


x2 + 3y + y2 = 5 at (1,1)


Here we have to differentiate the above equation with respect to x.


(x2 + 3y + y2) = (5)


(x2) + (3y) + (y2) = (5)


(xn) = n.xn – 1


2x + 3 + 2y = 0


2x + (3 + 2y) = 0


(3 + 2y) = – 2x



The Slope of the tangent at (1,1)is





The Slope of the tangent at (1,1) is


The Slope of the normal =


The Slope of the normal =


The Slope of the normal =


The Slope of the normal =


1