Equations of Normal and Tangent to a Curve

To find the equation of the tangent of the curve
$x^2+3xy+y^2=11$
at the point
$(x_1,y_1)=(1,2)$
$\frac{dy}{dx}$
/
$\frac{d}{dx}(x^2)+ \frac{d}{dx}(3xy)+ \frac{d}{dx}(y^2)= \frac{d}{dx}(11)$

$2x+3y+3x \frac{dy}{dx}+ 2y \frac{dy}{dx}=0$

$\frac{dy}{dx}(3x+2y)=-2x-3y$

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

The gradient of the curve at the point
$(2,1)$
is
$m= \frac{dy}{dx} |_{(2,1)} = \frac{-2 \times 2-3 \times 1}{3 \times 2 +2 \times 1}= - \frac{7}{8}$

Now use
$y-y_1=m(x-x_1)$
.
$y-1=- \frac{7}{8}(x-2) \rightarrow y=- \frac{7}{8}x+ \frac{11}{4}$
.
To find the equation of the normal use
$y-y_1=- \frac{1}{m}(x-x_1)$
.
$y-1= \frac{8}{7}(x-2) \rightarrow y= \frac{8}{7}x- \frac{9}{7}$
.