Question

The equation of the tangent curves x2 - 4x - 2y - 1 = 0 at the point (1, -2) is ...

Answer 1

Tangent
X² - 4x - 2y -1 = 0
or
2y = x² - 4x - 1
Y = 1/2 x² - 2x - 1/2
.The tangent gradient  (1, -2) -> m = y '= x - 2
For x = 1 -> m, = 1-2 = -1

y = m (x- x1) + y1
Y = - (x - 1) - 2
Y = - x + 1 -2
Y = - x - 1
or
X + y + 1 = 0

Answer 2

hi friend ,

given curve x²-4x-2y-1=0

→2y=x²-4x-1

differentiating with respect to x,we get

→2 dy/dx= 2x-4

→dy/dx =x-2

slope at the point (1,-2) is -1

equation of the tangent is

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

→(y+2)=-1(x-1)

→y+2=1-x

→x+y+1=0 is the equation of tangent ..


I hope this will help u :)