Step 1 We are given the density of the liquid as $\rho = 900\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$, the equation of elliptical plate as $4{y^2} + {x^2} = 1$, the major axis as $a = {\rm{1}}\;{\rm{m}}$, and the minor axis as $b = 0.5\;{\rm{m}}$.
We are asked to determine resultant force exerted on the plate and the location of the center of pressure from the $x$-axis.
Step 2 We will draw a free body diagram of the tank plate.
Here, $d{F_R}$ is the resultant force acting on the small element perpendicular to the plate, $p$ is the pressure acting on the plate, and $\overline y = y$ is the centroid of the element along $y$-axis.
Step 3 We will rewrite the elliptical equation.
\[\begin{array}{c} 4{y^2} + {x^2} = 1\\ {x^2} = 1 - 4{y^2}\\ x = \sqrt {1 - 4{y^2}} \end{array}\]
Step 4 We will find the height of the element from the top of the tank.
\[\begin{array}{l} y' = b - \overline y \\ y' = b - y \end{array}\]
Substitute the given value in the above equation.
\[\begin{array}{l} y' = \left( {{\rm{0}}{\rm{.5}}\;{\rm{m}}} \right) - y\\ y' = \left( {{\rm{0}}{\rm{.5}} - y} \right)\;{\rm{m}} \end{array}\]
Step 5 We will find the fluid pressure acting on the element.
\[\begin{array}{l} p = \rho gy'\\ p = \rho g\left( {0.5 - y} \right){\rm{m}} \end{array}\]
Here, $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ is the gravitational acceleration.
Substitute the given value in the above equation.
\[\begin{array}{c} p = \left( {900\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.5 - y} \right){\rm{m}}\\ = 8829\left( {0.5 - y} \right)\;{\rm{kg/m}} \cdot {{\rm{s}}^{\rm{2}}} \end{array}\]
Step 6 We will find the resultant force acting on the element.
\[\begin{array}{l} d{F_R} = dV\\ d{F_R} = pdA\\ d{F_R} = p\left( {2xdy} \right) \end{array}\]
Here, $2x$ is the width of the element, and $dy$ is the thickness of the element.
Substitute the given value in the above equation.
\[\begin{array}{l} d{F_R} = 8829\left( {0.5 - y} \right)\;\left( {2xdy} \right)\\ d{F_R} = 8829\left( {0.5 - y} \right)\;\left( {2\sqrt {1 - 4{y^2}} dy} \right)\\ d{F_R} = 17658\sqrt {1 - 4{y^2}} \left( {0.5 - y} \right)\;dy \end{array}\]
Step 7 We will find the resultant force exerted on the plate due to fluid.
\[{F_R} = \int\limits_{ - b}^b {d{F_R}} \]
Substitute the given value in the above equation.
\[\begin{array}{c} {F_R} = \int\limits_{ - 0.5}^{0.5} {17658\sqrt {1 - 4{y^2}} \left( {0.5 - y} \right)\;dy} \;{\rm{N}}\\ = 17658\int\limits_{ - 0.5}^{0.5} {\left( {0.5 - y} \right)\sqrt {1 - 4{y^2}} \;dy} \;{\rm{N}}\\ = 17658\left[ {\frac{{{{\sin }^{ - 1}}\left( {2y} \right)}}{8} + \frac{{{{\left( {1 - 4{y^2}} \right)}^{\frac{3}{2}}}}}{{12}} + \frac{{y\sqrt {1 - 4{y^2}} }}{4}} \right]_{ - 0.5}^{0.5}\;{\rm{N}}\\ = 17658\left[ \begin{array}{l} \left( {\frac{{{{\sin }^{ - 1}}\left( {2\left( {0.5} \right)} \right)}}{8} + \frac{{{{\left( {1 - 4{{\left( {0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{12}} + \frac{{y\sqrt {1 - 4{{\left( {0.5} \right)}^2}} }}{4}} \right) - \\ \left( {\frac{{{{\sin }^{ - 1}}\left( {2\left( { - 0.5} \right)} \right)}}{8} + \frac{{{{\left( {1 - 4{{\left( { - 0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{12}} + \frac{{y\sqrt {1 - 4{{\left( { - 0.5} \right)}^2}} }}{4}} \right) \end{array} \right]\;{\rm{N}} \end{array}\]
Solve the above equation.
\[\begin{array}{c} {F_R} = 17658\left[ \begin{array}{l} \left( {\frac{{{{\sin }^{ - 1}}\left( 1 \right)}}{8} + \frac{{{{\left( 0 \right)}^{\frac{3}{2}}}}}{{12}} + \frac{{y\sqrt 0 }}{4}} \right) - \\ \left( {\frac{{{{\sin }^{ - 1}}\left( { - 1} \right)}}{8} + \frac{{{{\left( 0 \right)}^{\frac{3}{2}}}}}{{12}} + \frac{{y\sqrt 0 }}{4}} \right) \end{array} \right]\;{\rm{N}}\\ = 6934.28\;{\rm{N}} \end{array}\]
Step 8 We will find the location of the resultant force.
\[\begin{array}{l} \overline y \int\limits_{ - b}^b {d{F_R}} = \int\limits_{ - b}^b {yd{F_R}} \\ \overline y {F_R} = \int\limits_{ - b}^b {yd{F_R}} \end{array}\]
Substitute the given value in the above equation.
\[\begin{array}{c} \overline y \left( {6934.28} \right) = \int\limits_{ - 0.5}^{0.5} {y\left( {17658} \right)\sqrt {1 - 4{y^2}} \left( {0.5 - y} \right)\;dy} \\ \overline y = \frac{{17658}}{{6934.28}}\int\limits_{ - 0.5}^{0.5} {y\sqrt {1 - 4{y^2}} \left( {0.5 - y} \right)\;dy} \\ = \frac{{17658}}{{6934.28}}\int\limits_{ - 0.5}^{0.5} {y\sqrt {1 - 4{y^2}} \left( {0.5 - y} \right)\;dy} \\ = \frac{{17658}}{{6934.28}}\left[ { - \frac{{{{\sin }^{ - 1}}\left( {2y} \right)}}{{64}} + \frac{{y{{\left( {1 - 4{y^2}} \right)}^{\frac{3}{2}}}}}{{16}} - \frac{{{{\left( {1 - 4{y^2}} \right)}^{\frac{3}{2}}}}}{{24}} - \frac{{y{{\left( {1 - 4{y^2}} \right)}^{\frac{1}{2}}}}}{{32}}} \right]_{ - 0.5}^{0.5}\\ = \frac{{17658}}{{6934.28}}\left[ \begin{array}{l} \left( \begin{array}{l} - \frac{{{{\sin }^{ - 1}}\left( {2\left( {0.5} \right)} \right)}}{{64}} + \frac{{\left( {0.5} \right){{\left( {1 - 4{{\left( {0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{16}} - \\ \frac{{{{\left( {1 - 4{{\left( {0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{24}} - \frac{{\left( {0.5} \right){{\left( {1 - 4{{\left( {0.5} \right)}^2}} \right)}^{\frac{1}{2}}}}}{{32}} \end{array} \right)\\ - \left( \begin{array}{l} - \frac{{{{\sin }^{ - 1}}\left( {2\left( { - 0.5} \right)} \right)}}{{64}} + \frac{{\left( { - 0.5} \right){{\left( {1 - 4{{\left( { - 0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{16}} - \\ \frac{{{{\left( {1 - 4{{\left( { - 0.5} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{24}} - \frac{{\left( { - 0.5} \right){{\left( {1 - 4{{\left( { - 0.5} \right)}^2}} \right)}^{\frac{1}{2}}}}}{{32}} \end{array} \right) \end{array} \right] \end{array}\]
Solve the above equation.
\[\begin{array}{c} \overline y = \frac{{17658}}{{6934.28}}\left[ \begin{array}{l} \left( { - \frac{{{{\sin }^{ - 1}}\left( 1 \right)}}{{64}} + \frac{{\left( {0.5} \right){{\left( 0 \right)}^{\frac{3}{2}}}}}{{16}} - \frac{{{{\left( 0 \right)}^{\frac{3}{2}}}}}{{24}} - \frac{{\left( {0.5} \right){{\left( 0 \right)}^{\frac{1}{2}}}}}{{32}}} \right)\\ - \left( { - \frac{{{{\sin }^{ - 1}}\left( { - 1} \right)}}{{64}} + \frac{{\left( { - 0.5} \right){{\left( 0 \right)}^{\frac{3}{2}}}}}{{16}} - \frac{{{{\left( 0 \right)}^{\frac{3}{2}}}}}{{24}} - \frac{{\left( { - 0.5} \right){{\left( 0 \right)}^{\frac{1}{2}}}}}{{32}}} \right) \end{array} \right]\\ \approx - 0.12\;{\rm{m}} \end{array}\]
So, the location of the center of pressure from the $x$-axis is $ - 0.12\;{\rm{m}}$.
