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 =