How do you find an equation of the line tangent to the curve at the given point x^3+y^3=36xy given point (18,18)?

Oct 21, 2016

$3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 36 \left(x \left(1\right) \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(1\right)\right)$

Simplify

$3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 36 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)$

Distribute

$3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 36 x \frac{\mathrm{dy}}{\mathrm{dx}} + 36 y$

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 36 x \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} + 36 y$

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 36 x\right) = - 3 {x}^{2} + 36 y$

Divide to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\frac{\mathrm{dy}}{\mathrm{dx}} \cancel{3 {y}^{2} - 36 x}}{\cancel{\left(3 {y}^{2} - 36 x\right)}} = \frac{- 3 {x}^{2} + 36 y}{3 {y}^{2} - 36 x}$

Simplify

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2} + 36 y}{3 {y}^{2} - 36 x}$

Factor out a 3

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(- {x}^{2} + 12 y\right)}{3 \left({y}^{2} - 12 x\right)}$

Simplify

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cancel{3} \left(- {x}^{2} + 12 y\right)}{\cancel{3} \left({y}^{2} - 12 x\right)}$

Simplify

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 y - {x}^{2}}{{y}^{2} - 12 x}$

Substitute in the point (18,18) to find the slope

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 \left(18\right) - {18}^{2}}{{18}^{2} - 12 \left(18\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

Now we have the slope(m), -1, and a point (18,18)

Use the point slope formula, $y = m \left(x - {x}_{1}\right) + {y}_{1}$

$y = - 1 \left(x - 18\right) + 18$

Simplify

$y = - x + 18 + 18$

$y = - x + 36$