A square is inscribed in the ellipse x^2 + 4y^2 = 4, with two side parallel to the x-axis. What is the area of the square.
+26367 A square is inscribed in the ellipse \(x^2 + 4y^2 = 4\), with two side parallel to the x-axis.
What is the area of the square.
\(\text{Let the side of the square $s$}\)
\(\begin{array}{|rcll|} \hline x^2 + 4y^2 &=& 4 \quad | \quad x=y=\dfrac{s}{2} \\ \left(\dfrac{s}{2}\right)^2 + 4*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ 5*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ \left(\dfrac{s}{2}\right)^2 &=& \dfrac{4}{5} \\ \dfrac{s^2}{4} &=& \dfrac{4}{5} \\ \mathbf{s^2} &=& \mathbf{\dfrac{16}{5}} \\ \mathbf{s^2} &=& \mathbf{3.2} \\ \hline \end{array}\)
The area of the square is \(\mathbf{3.2}\)
