Enjoying conics with integration!

Let's consider $P=(t,t^2), \hspace{5pt} t \in \mathbb{R}$ a generic point lying on $y=x^2$ . The line tangent to the parabola and passing from $P$ has equation $y_t=mx+q_t$ , where $m=2t$ is the slope of the parabola in $P$ . Let's impose that the line pass through $P$

$\displaystyle t^2=2t \cdot t +q_t \hspace{3pt} \Longrightarrow \hspace{3pt} q_t=-t^2 \hspace{3pt} \Longrightarrow \hspace{3pt} y_t=2tx-t^2$

It's well-known that the if a line has slope $m$ , than the slope of the normal is $-\frac{1}{m}$ . So, the normal to $y_n$ is $y_n=-\frac{x}{2t}+q_n$ . As done before, we impose the passage through $P$ to find $q_n$ , so

$\displaystyle y_n=-\frac{x}{2t}+t^2+\frac{1}{2}$

The normal line intersect the parabola at $Q$ , so we solve the system

$\displaystyle \begin{cases} \displaystyle y=-\frac{x}{2t}+t^2+\frac{1}{2} \\[5pt] \displaystyle y=x^2 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} x^2=-\frac{x}{2t}+t^2+\frac{1}{2} \hspace{5pt} \Longrightarrow \hspace{5pt} \displaystyle x_1=t, \hspace{5pt} x_2=\frac{-2t^2-1}{2t}$

where $x_1$ is exactly the $x-$ coordinate of $P$ and $x_2$ is the $x-$ coordinate of $Q$ . Now, we can evaluate the area $A$ between $y_n$ and the parabola

$\displaystyle A=\int_{x_1}^{x_2} y_n-x^2 \hspace{3pt} dx = \int_{t}^{\frac{-2t^2-1}{2t}} -\frac{x}{2t}+t^2+\frac{1}{2}-x^2 \hspace{3pt} dx=-\frac{(4 t^2 + 1)^3}{48 t^3}$

And, finally, we can solve the minimum problem

$\displaystyle min(A) \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{dA}{dt}=0 \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{d}{dt} \bigg(-\frac{(4 t^2 + 1)^3}{48 t^3}\bigg)=0 \hspace{5pt} \Longrightarrow \hspace{5pt} \frac{1}{16 x^4} - 4 x^2 + \frac{1}{4 x^2} - 1=0 \hspace{5pt} \Longrightarrow \hspace{5pt} x_1=\frac{1}{2}, \hspace{4pt} x_2=-\frac{1}{2}, \hspace{4pt} x_3=\frac{i}{2}, \hspace{4pt} x_4=-\frac{i}{2}$

$\displaystyle A\bigg(t=x_2=-\frac{1}{2}\bigg)=\boxed{\frac{4}{3}}$ .