# How do you find an equation for the tangent line to x^4=y^2+x^2 at (2, sqrt12)?

Jan 18, 2017

$y = \frac{7}{\sqrt{3}} x - \frac{8}{\sqrt{3}}$ or $\sqrt{3} y = 7 x - 8$

#### Explanation:

${x}^{4} = {y}^{2} + {x}^{2}$
${y}^{2} = {x}^{4} - {x}^{2}$
$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 4 {x}^{3} - 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3} - 2 x}{2 y}$

at $\left(2 , \sqrt{12}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \left({2}^{3}\right) - 2 \left(2\right)}{2 \sqrt{12}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{32 - 4}{4 \sqrt{3}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{28}{4 \sqrt{3}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{\sqrt{3}}$

The equation of tangent, $y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m = \frac{7}{\sqrt{3}} , {y}_{1} = \sqrt{12} \mathmr{and} {x}_{1} = 2$

$y - \sqrt{12} = \frac{7}{\sqrt{3}} \left(x - 2\right)$
$y - \sqrt{12} = \frac{7}{\sqrt{3}} x - \frac{14}{\sqrt{3}}$
$y = \frac{7}{\sqrt{3}} x - \frac{14}{\sqrt{3}} + \sqrt{12}$
$y = \frac{7}{\sqrt{3}} x - \frac{14}{\sqrt{3}} + \frac{\sqrt{36}}{\sqrt{3}}$
$y = \frac{7}{\sqrt{3}} x - \frac{14}{\sqrt{3}} + \frac{6}{\sqrt{3}}$
$y = \frac{7}{\sqrt{3}} x - \frac{8}{\sqrt{3}}$ or
$\sqrt{3} y = 7 x - 8$

Jan 18, 2017

$7 x - \sqrt{3} y - 8 = 0$. See tangent-inclusive graph. The graph is not to scale. There is contraction in the y-direction.

#### Explanation:

Differentiating,

$4 {x}^{3} = 2 x + 2 y y '$, giving $y ' = \frac{7}{\sqrt{3}}$, at $P \left(2 , \sqrt{12}\right)$.

The equation to the tangent at P is

y-sqrt13=7/sqrt3(x-2), giving

$7 x - \sqrt{3} y - 8 = 0$.

graph{(x^2-sqrt(x^2+y^2))(7x-sqrt3 y-8.2)=0 [-7, 7, -35, 35]}