The equation of the tangent to the curve...

The equation of the tangent to the curve given by `x ^(2) + 2x - 3y + 3 =0` at the point (1,2) is

A

`4x - 3y - 2 =0`

B

`3y - 4x - 2 =0`

C

`4x + 3y + 2 =0`

D

`4x + 3y -2=0`

Verified by Experts

Similar Questions

Explore conceptually related problems

The equation of the normal to the curve given by x ^(2) + 2x - 3y + 3 = 0 at the point (1,2) is

The equation of the tangent to the curve y = x^(3) - 6x + 5 at (2, 1) is

Find the equation of the tangent to the curve y=3x^2 at (1,1)

The slope of tangent line to the curve 4x^(2) +2xy +y^(2) =12 at the point (1,2) is

Slope of the tangent to the curve y=3x^2+2sinx at x=0 is

Find the equation of the tangent to the curve x^(2/3)+y^(2/3)=2 at (1,1).

Find the equation of the tangent to the curve y=(x-7)/((x-2)(x-3)) at the point where it cuts the x -axis.

The slope of the tangent to the curve y=x^3-1 at x=2 is………

The equation of the tangent to y=-2x^(2)+3 at x=1 is