\[x^2+3xy+y^2=11\]
at the point \[(x_1,y_1)=(1,2)\]
first find the gradient function \[\frac{dy}{dx}\]
/\[\frac{d}{dx}(x^2)+ \frac{d}{dx}(3xy)+ \frac{d}{dx}(y^2)= \frac{d}{dx}(11)\]
\[2x+3y+3x \frac{dy}{dx}+ 2y \frac{dy}{dx}=0\]
\[\frac{dy}{dx}(3x+2y)=-2x-3y\]
\[\frac{dy}{dx}= \frac{-2x-3y}{3x+2y}\]
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
\[y-y_1=- \frac{1}{m}(x-x_1)\]
.\[y-1= \frac{8}{7}(x-2) \rightarrow y= \frac{8}{7}x- \frac{9}{7}\]
.