Question

How do you find the equation of the tangent and normal line to the curve `y^3-3x-2y+6=0` at (9,3)?

Answer 1

Equation of the tangent: `y = 3/25x + 48/25`
Equation of the normal: `y = -25/3x + 78`

Explanation:

Start by finding the derivative.

`y^3 - 3x - 2y + 6 = 0`

`d/dx(y^3 - 3x - 2y + 6) = d/dx(0)`

`d/dx(y^3) + d/dx(-3x) + d/dx(-2y) + d/dx(6) = d/dx(0)`

`3y^2(dy/dx) - 3 - 2(dy/dx) + 0 = 0`

`dy/dx(3y^2 - 2) = 3`

`dy/dx= 3/(3y^2 -2)`

The slope of the tangent is given by substituting `(x, y)` into the derivative.

`m_"tangent" = 3/(3 xx 3^2 - 2)`

`m_"tangent" = 3/25`

Hence, the equation of the tangent, by point-slope form, is:

`y - y_1 = m(x- x_1)`

`y - 3 = 3/25(x - 9)`

`y - 3 = 3/25x - 27/25`

`y = 3/25x + 48/25`

The equation of the normal line is simple: it is perpendicular to the tangent. A line perpendicular to another has a slope that is the negative reciprocal of the original line.

Thus, the slope of the normal is `-25/3`.

`y - y_1 = m(x- x_1)`

`y - 3 = -25/3(x - 9)`

`y - 3 = -25/3x + 75`

`y = -25/3x + 78`

Hopefully this helps!