Orthocenter Locus

Here is a solution based on dot product.

Let $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ be the ellipse in the general case. Then, $A=\left( -a,0 \right)$ , $B=\left( a,0 \right)$ . Let $C=\left( {{x}_{0}},{{y}_{0}} \right)$ . Denote by $P$ the orthocenter.

Since $PC$ is perpendicular to the $x$ -axis, $P$ and $C$ have the same $x$ -coordinate, i.e. $P=\left( {{x}_{0}},t \right)$ . Now, $P$ is the orthocenter iff

$\begin{aligned} \overrightarrow{AC}\bot \overrightarrow{BP} & \Leftrightarrow \overrightarrow{AC}\cdot \overrightarrow{BP}=0 \\ & \Leftrightarrow \left( {{x}_{0}}+a,{{y}_{0}} \right)\cdot \left( {{x}_{0}}-a,t \right)=0 \\ & \Leftrightarrow \left( {{x}_{0}}+a \right)\left( {{x}_{0}}-a \right)+{{y}_{0}}t=0 \\ & {{y}_{0}}t={{a}^{2}}-{{x}_{0}}^{2} \\ \end{aligned}$ Since ${{x}_{0}}\ne \pm a$ , we get $t\ne 0$ , thus, ${{y}_{0}}=\frac{{{a}^{2}}-{{x}_{0}}^{2}}{t}$ Point $C$ belongs to the ellipse, hence

$\begin{aligned} \frac{{{x}_{0}}^{2}}{{{a}^{2}}}+\frac{{{y}_{0}}^{2}}{{{b}^{2}}}=1 & \Leftrightarrow \frac{{{x}_{0}}^{2}}{{{a}^{2}}}+\frac{{{\left( \frac{{{a}^{2}}-{{x}_{0}}^{2}}{t} \right)}^{2}}}{{{b}^{2}}}=1 \\ & \Leftrightarrow \frac{{{\left( {{a}^{2}}-{{x}_{0}}^{2} \right)}^{2}}}{{{b}^{2}}{{t}^{2}}}=\frac{{{a}^{2}}-{{x}_{0}}^{2}}{{{a}^{2}}} \\ & \Leftrightarrow \frac{{{a}^{2}}-{{x}_{0}}^{2}}{{{b}^{2}}}=\frac{{{t}^{2}}}{{{a}^{2}}} \\ & \Leftrightarrow \frac{{{x}_{0}}^{2}}{{{a}^{2}}}+\frac{{{t}^{2}}}{{{\left( \frac{{{a}^{2}}}{b} \right)}^{2}}}=1 \\ \end{aligned}$ This means that the locus is the ellipse with equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \frac{{{a}^{2}}}{b} \right)}^{2}}}=1$ , with the exception of its co-vertices, which are points $A$ and $B$ . Its semi-major and semi-minor axes are ${a}'=\dfrac{{{a}^{2}}}{b} \ \ \ \ \ and \ \ \ \ \ {b}'=a$ respectively.

Thus, the area in question is $A={a}'{b}'\pi =\dfrac{{{a}^{3}}}{b}\pi$ Taking $a=1$ and $b=\dfrac{1}{2}$ we find $x=\boxed{2}$ .

We'll do the solution for a general ellipse having a horizontal semi-major axis of $a$ and vertical semi-minor axis of $b$ . Then point $A = (-a, 0) , B = (a, 0)$ and $C = ( a \cos t, b \sin t )$ where $t$ is arbitrary.

Drop a vertical perpendicular from $C$ to $AB$ , then its equation is $x = a \cos t$

Next drop a perpendicular from $A$ to $BC$ :

$BC$ normal is the perpendicular vector to $( a - a \cos t, - b \sin t )$ , so the normal is $(b \sin t, a - a \cos t )$

the parametric equation of the perpendicular line is $p(s) = (-a, 0) + s (b \sin t, a - a \cos t ) = (-a + s b \sin t , s a (1 - \cos t ) )$

since $x = a \cos t$ , then $-a + s b \sin t = a \cos t$

hence $s = \dfrac{ a (\cos t + 1)} {b \sin t}$

substituting this value of $s$ in the equation of the perpendicular line results in,

$y = a^2 \dfrac{(1 + \cos t )(1 - \cos t )}{b \sin t } = \dfrac{a^2 (1 - \cos^2 t)}{b \sin t } = \dfrac{ a^2 }{b} \sin t$

Hence the orthocenter at $t$ is $( a \cos t , \dfrac{a^2}{b} \sin t )$ and this parametric equation is for an ellipse with a horizontal semi-minor axis of length $a$ and a vertical semi-major axis of length $\dfrac{a^2}{b}$ , and a has an enclosed area of $\dfrac{a^3}{b} \pi$ .

Plugging in $a = 1 , b = \dfrac{1}{2}$ , yields an area of $2 \pi$ , thus $x= \boxed{2}$