Given the equation below;
[tex](x^2+4)y=13[/tex]To find the derivative of the equation above, we will defferentiate implicitly, this is shown below
[tex]\begin{gathered} (x^2+4)y=13 \\ \text{ Remove bracket} \\ x^2(y)+4(y)=13 \\ x^2y+4y=13 \end{gathered}[/tex][tex]\begin{gathered} 2xy+x^2\frac{\mathrm{d}y}{dx}+4\frac{\mathrm{d}y}{dx}=0 \\ x^2\frac{\mathrm{d}y}{dx}+4\frac{\mathrm{d}y}{dx}=-2xy \\ \text{factor }\frac{\mathrm{d}y}{dx} \\ \frac{\mathrm{d}y}{dx}(x^2+4)=-2xy \end{gathered}[/tex][tex]\frac{\mathrm{d}y}{dx}=\frac{-2xy}{(x^2+4)}[/tex]At point (3, 1) we can obtain the value of the slope ( dy/dx), by substituting for x and y.
[tex]\begin{gathered} x=3,y=1 \\ \frac{\mathrm{d}y}{dx}=\frac{-2xy}{(x^2+4)} \\ \frac{\mathrm{d}y}{dx}=\frac{-2(3)(1)}{3^2+4}=\frac{-6}{9+4}=-\frac{6}{13} \end{gathered}[/tex]Using the point-slope equation of a line formula as shown below
[tex]y-y_1=m(x-x_1)[/tex]We can use the above to find the tangent to the line equation. then we have:
[tex]\begin{gathered} y-1=-\frac{6}{13}(x-3) \\ y-1=-\frac{6}{13}x+\frac{18}{13} \\ y=-\frac{6}{13}x+\frac{18}{13}+1 \\ y=-\frac{6}{13}x+\frac{31}{13} \end{gathered}[/tex]To find the normal line equation, we have to use the condition for perpendicularity of two lines, that is
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ If_{} \\ m_1\Rightarrow\text{slope of the tangent line} \\ m_2\Rightarrow\text{slope of the normal line} \\ m_2=-\frac{1}{m_1} \\ \text{Thus, the slope of the normal line is} \\ =\frac{13}{6} \end{gathered}[/tex]Using point-slope equation of a line we can also find the normal line equation.
[tex]\begin{gathered} y-1=\frac{13}{6}(x-3) \\ y-1=\frac{13}{6}x-\frac{13}{2} \\ y=\frac{13}{6}x-\frac{13}{2}+1 \\ y=\frac{13}{6}x-\frac{11}{2} \end{gathered}[/tex]Hence, the tangent line and normal line equation are
[tex]\begin{gathered} \text{tangent line}\Rightarrow y=-\frac{6}{13}x+\frac{31}{13}\text{ } \\ \text{normal line}\Rightarrow y=\frac{13}{6}x-\frac{11}{2} \end{gathered}[/tex]This can be simplified as;
[tex]\begin{gathered} \text{tangent line}\Rightarrow y=-\frac{6}{13}x+2\frac{5}{13}\text{ } \\ \text{normal line}\Rightarrow y=2\frac{1}{6}x-5\frac{1}{2} \end{gathered}[/tex]