Answer
tangent line:
$y = -\frac{4}{5}x+\frac{13}{5}$
normal line:
$y = \frac{5}{4}x-\frac{3}{2}$
Work Step by Step
$x^2+4xy+y^2=13$
We can use implicit differentiation to find $\frac{dy}{dx}$:
$2x+4y+4x\frac{dy}{dx}+2y\frac{dy}{dx}=0$
$4x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-4y$
$\frac{dy}{dx}=-\frac{2x+4y}{4x+2y}$
We can find $\frac{dy}{dx}$ when $x = 2$ and $y=1$:
$\frac{dy}{dx}=-\frac{2(2)+4(1)}{4(2)+2(1)}$
$\frac{dy}{dx}=-\frac{8}{10}$
$\frac{dy}{dx}=-\frac{4}{5}$
The tangent line has a slope of $-\frac{4}{5}$ at this point.
We can find the equation of the tangent line:
$y-1 = (-\frac{4}{5})(x-2)$
$y-1 = -\frac{4}{5}x+\frac{8}{5}$
$y = -\frac{4}{5}x+\frac{13}{5}$
The normal line has a slope of $\frac{5}{4}$ at this point.
We can find the equation of the normal line:
$y-1 = (\frac{5}{4})(x-2)$
$y-1 = \frac{5}{4}x-\frac{5}{2}$
$y = \frac{5}{4}x-\frac{3}{2}$
