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\title{Smallest Area of a Triangle Formed from the Tangent Line of a Parabola}
\author{Roop Pal}
\maketitle
\section*{Introduction}
This problem is an applied optimization problem. The problem is to minimize the area of the triangle formed by a tangent line to the function $y=1-\frac{1}{9}x^2$. The triangle is defined by the origin, the x-intercept of the tangent line, and the y-intercept of the tangent line. Only triangles formed in the first quadrant are of concern.
\section*{Variables}
The original parabola is represented by the function $f$. The relevant variables are the area, represented by $A$; the x-intercept, represented by $i$, and the y-intercept, represented by $j$. The x-coordinate of the tangency point is represented by $t$, so the y-coordinate of the tangency point is represented by $f(t)=1-\frac{1}{9}t^2$. The slope of the tangent line, then, is represented by $f'(t)=-\frac{2}{9}t$. The x-intercept of $f$ is evaluated below.
\begin{center}
$0=1-\frac{1}{9}x^2$ \\
$x^2=9$ \\
$x=\pm 3$ \\
$x=3$ as we are focused on the first quadrant.
\end{center}
A diagram is drawn below.
\begin{center}
\includegraphics[scale=0.75]{Figure1.png}
\text{Made with MS Paint}
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\section*{Objective Function and Range}
First, we must find the objective function $A(t)$ and range constraining $t$.
\subsection*{Objective Function}
The objective function is $A(t)$, or the area of the triangle as dependent on $t$, the arbitrary x-coordinate on the function $f$. $A(t)=\frac{1}{2}\cdot i j$ by the formula for the area of a triangle. The tangent line in slope-intercept form is
\begin{center}
$y-f(t)=f'(t)(x-t)$ by the definition of a tangent line. \\
$y-(1-\frac{1}{9}t^2)=-\frac{2}{9}t(x-t)$ by substitution. \\
$y-1+\frac{1}{9}t^2=-\frac{2}{9}tx+\frac{2}{9}t^2$ \\
$y=-\frac{2}{9}tx+\frac{1}{9}t^2+1$ \\
Now we must evaluate for the intercepts $i$ and $j$. \\
$0=-\frac{2}{9}ti+\frac{1}{9}t^2+1$ \\
$\frac{2}{9}ti=\frac{1}{9}t^2+1$ \\
$i=\frac{9}{2t}\cdot\frac{t^2}{9}+\frac{9}{2t}$ \\
$i=\frac{t}{2}+\frac{9}{2t}$ \\
\hfill \linebreak
$j=-\frac{2}{9}t\cdot 0+\frac{1}{9}t^2+1$ \\
$j=\frac{1}{9}t^2+1$ \\
Now that we have $i$ and $j$ in terms of $t$, we can use them to find the area function or objective function $A(t)$. \\
$A(t)=\frac{1}{2}\cdot ij$ by the definition of the area of a triangle. \\
$A(t)=\frac{1}{2}\cdot(\frac{t}{2}+\frac{9}{2t})\cdot(\frac{1}{9}t^2+1)$ by substitution. \\
$A(t)=\frac{t^3}{36}+\frac{t}{2}+\frac{9}{4t}$ by expansion.
\end{center}
\subsection*{Range}
The range of $t$ is constrained to the first quadrant. It ranges from 0 to the x-intercept of $f$, which, as evaluated above, is 3. Thus, $t\in[0,3]$.
\section*{Optimization}
To optimize $A(t)$, we must first find $A'(t)$.
\begin{center}
$A'(t)=\frac{t^2}{12}+\frac{1}{2}-\frac{9}{4t^2}$ by the Power Rule. \\
Then, we must set the function to 0 to find points of interest. \\
$\frac{t^2}{12}+\frac{1}{2}-\frac{9}{4t^2}=0$ \\
$t^4+6t^2-27=0$ by multiplying by $12t^2$. \\
$(t^2+9)(t^2-3)=0$ \\
$t=\pm \sqrt{3}$ if we solve for $t$. \\
$t=\sqrt{3}$ since we are working in the first quadrant. \\
\end{center}
Now we must analyze this critical point and the endpoints. The points are 0, $\sqrt{3}$, and 3.
\begin{center}
$A(0)=\frac{9}{4\cdot 0}$ \\
$A(0)=\infty$ \\
\hfill \linebreak
$A(\sqrt{3})=\frac{\sqrt{3}^3}{36}+\frac{\sqrt{3}}{2}+\frac{9}{4 \sqrt{3}}$ \\
$A(\sqrt{3})=\frac{\sqrt{3}}{12}+\frac{6\sqrt{3}}{12}+\frac{9\sqrt{3}}{12}$ \\
$A(\sqrt{3})=\frac{16\sqrt{3}}{12}$ \\
$A(\sqrt{3})=\frac{4\sqrt{3}}{3}$ \\
\hfill \linebreak
$A(3)=\frac{3^3}{36}+\frac{3}{2}+\frac{9}{4\cdot 3}$ \\
$A(3)=\frac{3}{4}+\frac{6}{4}+\frac{3}{4}$ \\
$A(3)=\frac{12}{4}$ \\
$A(3)=3$
\end{center}
As seen, $t=\sqrt{3}$ yields the smallest area of $\frac{4\sqrt{3}}{3}$. Now we must calculate the intercepts $i$ and $j$, the y-coordinate of the tangent point $f(t)$, and the equation of the tangent line in point slope form $y-f(t)=f'(t)t(x-t)$.
\begin{center}
$i=\frac{t}{2}+\frac{9}{2t}$ \\
$i=\frac{\sqrt{3}}{2}+\frac{9}{2\sqrt{3}}$ \\
$i=\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}$ \\
$i=\frac{4\sqrt{3}}{2}$ \\
$i=2\sqrt{3}$ \\
\hfill \linebreak
$j=\frac{1}{9}t^2+1$ \\
$j=\frac{1}{3}+1$ \\
$j=\frac{4}{3}$ \\
\hfill \linebreak
$f(\sqrt{3})=1-\frac{1}{3}$ \\
$f(\sqrt{3})=\frac{2}{3}$ \\
\hfill \linebreak
$y-\frac{2}{3}=-\frac{2\sqrt{3}}{9}({x-\sqrt{3}})$
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\section*{Conclusion}
In conclusion, the triangle with the smallest area is the one with the following properties. The tangent point is $(\sqrt{3},\frac{2}{3})$. The equation of the tangent line is $y-\frac{2}{3}=-\frac{2\sqrt{3}}{9}({x-\sqrt{3}})$. The line's intercepts are $(2\sqrt{3},0)$ and $(0,\frac{4}{3})$. The area of the triangle is $\frac{4\sqrt{3}}{3}$.
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