If \(P\) is a point on an ellipse \(5 x^{2}+4 y^{2}=80\) whose foci are \(S\) and \(\mathrm{S}^{\prime}\). Then \(\mathrm{PS}+\mathrm{PS}^{\prime}=\ldots \ldots \ldots\) (a) \(4 \sqrt{5}\) (b) 4 (c) 8 (d) 10

#tag_title# Find the major and minor axes # #tag_content# From the standard form of the ellipse equation, we have the semi-major axis \(a = \sqrt{\frac{80}{4}} = \sqrt{20}\), and the semi-minor axis \(b = \sqrt{\frac{80}{5}} = \sqrt{16}\). #tag_title# Locate the foci # #tag_content# Since the given ellipse is elongated along the x-axis, we will use the formula \(c = \sqrt{a^2 - b^2}\) to find the distance from the center to the foci. In this case, \(c = \sqrt{(\sqrt{20})^2 - (\sqrt{16})^2} = \sqrt{4}\). Therefore, the foci are located at \(\mathrm{S}(\pm 2, 0)\) and \(\mathrm{S'}(\mp 2, 0)\). #tag_title# Find the sum of distances # #tag_content# The sum of distances from any point on the ellipse to its foci is always equal to the length of the major axis. In this ellipse, the major axis is equal to \(2\sqrt{20}\). Therefore, \(\mathrm{PS} + \mathrm{PS'} = 2\sqrt{20} = (4\sqrt{5})\). The correct answer is (a) \(4\sqrt{5}\).