Question

find equation of tangent and normal to the curve y=x(2-x) at point (2,0)​

Answer 1

Answer:

2x + y - 4 = 0

x - 2y - 2 = 0

Step-by-step explanation:

To find---> Equation of tangent and normal to the curve y = x ( 2 - x ) at point ( 2 , 0 ).

Solution---> ATQ,

y = x ( 2 - x )

y = 2x - x²

Differentiating with respect to x , we get,

=> dy / dx = d / dx ( 2x - x² )

=> dy / dx = 2 ( 1 ) - 2x

=> dy / dx = 2 - 2x

Slope of tangent at ( 2 , 0 ) = (dy / dx) at ( 2 , 0 )

= 2 - 2 ( 2 )

= 2 - 4

= - 2

Slope of normal = - ( dx / dy ) at ( 2 , 0 )

= - ( - 1 / 2 )

= 1 / 2

Equation of tangent at ( x₁ , y₁ )

( y - y₁ ) = dy / dx ( x - x₁ )

Equation of tangent at ( 2 , 0 )

y - 0 = - 2 ( x - 2 )

=> y = -2x + 4

=> 2x + y - 4 = 0

Equation of normal at ( x₁ , y₁ )

( y - y₁ ) = - ( dx / dy ) ( x - x₁ )

Equation of normal at ( 2 , 0 )

y - 0 = ( 1 / 2 ) ( x - 2 )

=> 2y = ( x - 2 )

=> x - 2y - 2 = 0