find the equation of the tangent line

Question

Answer 1

The given equation x²- 2xy - y²- 3y = -4 and the tangent  line at  the point  ( 2,1)

Differentiate on  each side

2x - 2y - 2x ( dy/dx) - dy/dx - 3 = 0

Substitute  x = 2 and y = 1 in above equation

2(2) - 2(1) - 2(2) dy/dx - dy/dx - 3 = 0

2 - 5 dy/dx - 3 = 0

dy/dx = -1/5

Hence the equation of  tangent  line

( y - y1)  = m ( x - x1)

Substitute  x = 2 and y = 1 in above equation

( y - 1)  = -1/5 ( x -2 )

5( y - 1) = -1 ( x - 2)

5y - 5 = - x +2

5y =  - x +7

y = ( - x +7)/5

The tangent line of the curve  y = ( - x +7)/5.

Answer 2

b) The given equation e^xy +xy = e^6 +6 and the tangent  line at  the point  ( 2, 3)

Differentiate on  each side

y +x dy/dx +e^xy (y+xdy/dx) = 0

Substitute  x = 2 and y = 3 in above equation

3+2dy/dx+ e^6 (3+2dy/dx) = 0

3+2dy/dx +3e^6+2dy/dx e^6 = 0

3(1+e^6) +2dy/dx(1+e^6) = 0

2dy/dx = -3

dy/dx = -3/2

Hence the equation of  tangent  line

( y - y1)  = m ( x - x1)

Substitute  x = 2 and y = 3 in above equation

( y - 3)  = -3/2 ( x -2 )

2( y -3) = -3 ( x - 2)

2y -6 = -3x+6

2y = -3x+12

y =( -3x+12)/2

The tangent line of the curve  y = 3(-x+4)/2.