Question

find
the Equation of tangent and normal to the curve
y = (x² +3x+2) at the point (1,-1)​

Answer 1

Answer---> 5x - y - 6 = 0

x + 5y + 4 = 0

To find---> The equation of tangent and normal to the curve y = ( x² + 3x + 2 ) at the point ( 1 , - 1 )

Solution---> y = ( x² + 3x + 2 )

Differentiating with respect to x

dy / dx = d / dx ( x² + 3x + 2 )

= d / dx ( x² ) + 3 d / dx ( x ) + d / dx ( 2 )

= 2x + 3 ( 1 ) + 0

dy / dx = 2x + 3

Slope of tangent at ( 1 , - 1 ) = 2 ( 1 ) + 3

= 2 + 3

= 5

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

= - ( 1 / 5 )

Equation of tangent at ( x₁ , y₁ )

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

Equation of tangent at ( 1 , - 1 ) .

{ y - ( - 1 ) } = 5 ( x - 1 )

=> y + 1 = 5x - 5

=> 5x - y - 5 - 1 = 0

=> 5x - y - 6 = 0

Equation of normal at ( x₁ , y₁ )

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

Equation of normal at ( 1 , - 1 )

{ y - ( - 1 ) } = - ( 1 / 5 ) ( x - 1 )

=> 5( y + 1 ) = - ( x - 1 )

=> 5y + 5 = -x + 1

=> x + 5y + 5 - 1 = 0

=> x + 5y + 4 = 0