Find The Equation of The Normal To The Curve

[KAISER91]

Homework Statement

Find the equation of the normal to the curve; 2x^3 - x(^2)y + y^3 = 1

at the point (2, -3)

Homework Equations

The Attempt at a Solution

So I used implicit differentiation to get the gradient of the tangent.

2x^3 - x(^2)y + y^3 = 1

dy/dx = (2xy - 6x^2) / (3y^2 - x^2) at (2,-3)

= -36 / 23 = Gradient = m


Now I use y - y1 = m(x -x1) to get the equation of the tangent.

I'm having trouble here, arranging the equation.

What am I doing wrong?

y - y1 = m(x -x1)
y + 3 = -36/23 (x - 2)
y + 3 = (-36/23)x + 72/23



Not sure how to arrange that.

The gradient and equation of the normal will be easier if I knew how to arrange this.

Thanks.

 

[rock.freak667]

The Attempt at a Solution

So I used implicit differentiation to get the gradient of the tangent.

2x^3 - x(^2)y + y^3 = 1

dy/dx = (2xy - 6x^2) / (3y^2 - x^2) at (2,-3)

= -36 / 23 = Gradient = m

Right, -36/23 is the gradient of the tangent the the curve. You want the gradient of the normal to the tangent.

gradient of tangent*gradient of normal = -1