Given:
[tex]x^2+2y^2-3x-4y+2=0[/tex]Let's find the equations of the tangent line and normal line at (1, 2)
• Tangent line:
To find the equation of the tangent line, let's find the first derivative of the equation:
[tex]\begin{gathered} \frac{d}{dx}(x^2+2y^2-3x-4y+2)=\frac{d}{dx}(0) \\ \\ 2x+4yy^{\prime}-3-4y^{\prime}=0 \end{gathered}[/tex]Now, rewrite the equation for y'
[tex]\begin{gathered} 4yy^{\prime}-4y^{\prime}=-2x+3 \\ \\ 4y^{\prime}(y-1)=-2x+3_{} \\ \\ 4y^{\prime}=-\frac{2x}{y-1}+\frac{3}{y-1} \\ \\ y^{\prime}=-\frac{2x}{4(y-1)}+\frac{3}{4(y-1)} \\ \\ \frac{dy}{dx}=\frac{x}{2(y-1)}+\frac{3}{4(y-1)} \end{gathered}[/tex]At the given point: (1, 2)
To find the tangential line, substitute 1 for x and 2 for y:
[tex]\begin{gathered} \frac{dy}{dx}=\frac{1}{2(2-1)}+\frac{3}{4(2-1)} \\ \\ \frac{dy}{dx}=\frac{1}{2}+\frac{3}{4} \\ \\ \frac{dy}{dx}=\frac{1}{4} \end{gathered}[/tex]Here, 1/4 is the slope.
Take the point slope form and input 1/4 for m and (x1, y1) = (1, 2):
[tex]\begin{gathered} y-y1=m(x-x1) \\ \\ y-2=\frac{1}{4}(x-1) \\ \\ y-2=\frac{1}{4}x-\frac{1}{4} \\ \\ y=\frac{1}{4}x-\frac{1}{4}+2 \\ \\ y=\frac{1}{4}x+\frac{7}{4} \end{gathered}[/tex]Therefore, the equation of the tangent line of the curve at the given point is:
[tex]y=\frac{1}{4}x+\frac{7}{4}[/tex]• Normal line:
The normal line is perpendicular to the tangent line.
The slope of perpendicular lines are negative reciprocals of each other.
Given that the slope of the tangent line is 1/4, let's find the slope of the normal line:
[tex]\begin{gathered} m_1m_2=-1 \\ \\ m_2=-\frac{1}{m_1} \\ \\ m_2=-\frac{1}{\frac{1}{4}} \\ \\ m_2=-4 \end{gathered}[/tex]The slope of the normal line is = -4
Input the points (1, 2) for x1 and y1, then -4 for m in the point slope form:
[tex]\begin{gathered} y-y1=m(x-x1) \\ \\ y-2=-4(x-1) \\ \\ y-2=-4x+4 \\ \\ y=-4x+4+2 \\ \\ y=-4x+6 \end{gathered}[/tex]Therefore, the equation of the normal line of the curve at the given point is:
[tex]y=-4x+6[/tex]ANSWER:
[tex]\begin{gathered} \text{ tangent line: y = }\frac{1}{4}x+\frac{7}{4} \\ \\ \text{ Normal line: y = -4x + 6} \end{gathered}[/tex]
