Question

find equation of tangent and normal if y=x^2 at (-1,1)
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Answer 1

Answer:

2x + y + 1 = 0

x - 2y + 3 = 0

Step-by-step explanation:

To find---> Equation of tangent and normal of

y = x² at ( -1 , 1 ) .

Solution---> ATQ,

y = x²

Differentiating with respect to x,

=> dy / dx = d / dx ( x² )

= 2x

Slope of tangent at ( - 1 , 1 ) = dy/dx at ( - 1 , 1 )

= 2 ( - 1 )

= - 2

Equation of tangent at ( x₁ , y₁ )

( y - y₁ ) = ( dy/dx ) at ( x₁ , y₁ ) ( x - x₁ )

Equation of tangent at ( - 1 , 1 )

( y - 1 ) = ( - 2 ) { x - ( - 1 ) }

=> ( y - 1 ) = - 2 ( x + 1 )

=> y - 1 = -2x - 2

=> 2x + y + 2 - 1 = 0

=> 2x + y + 1 = 0

Slope of normal = - ( dx / dy ) at ( - 1 , 1 )

= - 1 / ( - 2 )

= 1 / 2

Equation of normal at ( x₁ , y₁ )

( y - y₁ ) = - ( dx / dy ) at ( x₁ , y₁ ) ( x - x₁ )

Equation of normal at ( -1 , 1 ) ,

y - 1 = ( 1 / 2 ) { x - ( - 1 ) }

=> 2 ( y - 1 ) = ( x + 1 )

=> 2y - 2 = x + 1

=> - x + 2y - 2 - 1 = 0

=> - x + 2y - 3 = 0

=> x - 2y + 3 = 0