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

Sep 27, 2016

Equation of the tangent: $y = \frac{3}{25} x + \frac{48}{25}$
Equation of the normal: $y = - \frac{25}{3} x + 78$

#### Explanation:

Start by finding the derivative.

${y}^{3} - 3 x - 2 y + 6 = 0$

$\frac{d}{\mathrm{dx}} \left({y}^{3} - 3 x - 2 y + 6\right) = \frac{d}{\mathrm{dx}} \left(0\right)$

$\frac{d}{\mathrm{dx}} \left({y}^{3}\right) + \frac{d}{\mathrm{dx}} \left(- 3 x\right) + \frac{d}{\mathrm{dx}} \left(- 2 y\right) + \frac{d}{\mathrm{dx}} \left(6\right) = \frac{d}{\mathrm{dx}} \left(0\right)$

$3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 3 - 2 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 0 = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 2\right) = 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{3 {y}^{2} - 2}$

The slope of the tangent is given by substituting $\left(x , y\right)$ into the derivative.

${m}_{\text{tangent}} = \frac{3}{3 \times {3}^{2} - 2}$

${m}_{\text{tangent}} = \frac{3}{25}$

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

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 3 = \frac{3}{25} \left(x - 9\right)$

$y - 3 = \frac{3}{25} x - \frac{27}{25}$

$y = \frac{3}{25} x + \frac{48}{25}$

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 $- \frac{25}{3}$.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 3 = - \frac{25}{3} \left(x - 9\right)$

$y - 3 = - \frac{25}{3} x + 75$

$y = - \frac{25}{3} x + 78$

Hopefully this helps!