# How do you use implicit differentiation to find an equation of the tangent line to the curve x^2 + 2xy − y^2 + x = 39 at the given point (5, 9)?

Mar 3, 2018

$29 x - 8 y = 73$

#### Explanation:

Differentiate the equation with respect to $x$:

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

$2 x + 2 y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$\left(2 x - 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 1 - 2 x - 2 y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + 2 x + 2 y}{2 y - 2 x}$

The equation of the tangent line is:

$y = {y}_{0} + y ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

where ${x}_{0} = 5$, ${y}_{0} = 9$ and:

$y ' \left({x}_{0}\right) = \frac{1 + 10 + 18}{18 - 10} = \frac{29}{8}$

then:

$y = 9 + \frac{29}{8} \left(x - 5\right)$

$y = \frac{29}{8} x - \frac{73}{8}$

or:

$29 x - 8 y = 73$

$29 x - 8 y - 73 = 0$
is the equation of the tangent

#### Explanation:

Given:
${x}^{2} + 2 x y - {y}^{2} + x = 39$
when x=5, and y=9
${x}^{2} + 2 x y - {y}^{2} + x = {5}^{2} + 2 \times 5 \times 9 - {9}^{2} + 5$
$= 25 + 90 - 81 + 5 = \left(25 + 5\right) + \left(90 - 81\right)$
$= 30 + 9 = 39$
${x}^{2} + 2 x y - {y}^{2} + x = 39$
verified

We have
${x}^{2} + 2 x y - {y}^{2} + x = 39$
Differentiating both sides wrt x
$2 x + 2 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$
Substituting x=5 and y=9
$2 \times 5 + 2 \left(5 \times \frac{\mathrm{dy}}{\mathrm{dx}} + 9\right) - 2 \times 9 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$10 + 10 \frac{\mathrm{dy}}{\mathrm{dx}} + 18 - 18 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$
$\left(10 + 18 + 1\right) + \left(10 - 18\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\left(28 + 1\right) - 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$29 = 8 \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{29}{8}$
Slope of the tangent is
$m = \frac{29}{8}$
Passing through the point
$P \equiv \left(5 , 9\right)$
Equation of the tangent is
$\frac{y - 9}{x - 5} = \frac{29}{8}$

$8 \left(y - 9\right) = 29 \left(x - 5\right)$
$8 y - 72 = 29 x - 145$
$27 x - 8 y + 72 - 145 = 0$

$29 x - 8 y - 73 = 0$