Q1 of 149 Page 16

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 =


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