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  
\[(x_1,y_1)=(1,2)\]
  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  
\[y-y_1=- \frac{1}{m}(x-x_1)\]
.
\[y-1= \frac{8}{7}(x-2) \rightarrow y= \frac{8}{7}x- \frac{9}{7}\]
.

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