If \(\mathrm{P}(\mathrm{m}, \mathrm{n})\) is a point on an ellipse \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) with foci \(\mathrm{S}\) and \(\mathrm{S}^{\prime}\) and eccentricity e, then area of \(\mathrm{SPS}^{\prime}\) is \(\ldots \ldots \ldots\) (a) \(\mathrm{ae} \sqrt{\left(a^{2}-\mathrm{m}^{2}\right)}\) (b) \(\mathrm{ae} \sqrt{\left(b^{2}-\mathrm{m}^{2}\right)}\) (c) \(b e \sqrt{\left(b^{2}-m^{2}\right)}\) (d) be \(\sqrt{\left(a^{2}-m^{2}\right)}\)

Since the ellipse has its foci S and S' on the x-axis, the coordinates will be of the form S(-ae, 0) and S'(ae, 0), where a is the semi-major axis, and e is the eccentricity.

Using the distance formula, we can find the lengths of SP and S'P: SP = \(\sqrt{(m + ae)^2 + n^2}\) S'P = \(\sqrt{(m - ae)^2 + n^2}\)

For any point P on the ellipse, the sum of the distances from P to the foci S and S' is constant and equal to 2a: SP + S'P = 2a Substitute SP and S'P with the expressions from the previous step \(\sqrt{(m + ae)^2 + n^2} + \sqrt{(m - ae)^2 + n^2} = 2a\)

Using the half-base times height formula for the area of the triangle, we get: Area = (1/2) * SP * S'P * sin(∠SPS') Since the coordinates of S, S', and P have been found, it's possible to find the sin(∠SPS'). Let ∠SPM=x and ∠S'PM=y, then sin(∠SPS')=sin(x+y), sin(x+y)=sin(x)cos(y)+cos(x)sin(y) From the cosine law we have: \(\cos(x)=\frac{\left(m+ae\right)^2+n^{2}-m^{2}-n^{2}}{2*SP*ae}\) \(\cos(y)=\frac{\left(m-ae\right)^2+n^{2}-m^{2}-n^{2}}{2*S'P*ae}\) \(\sin(x)=\frac{n}{SP}\) \(\sin(y)=\frac{n}{S'P}\) Substitute these values into the equation of sin(∠SPS') sin(∠SPS')= \(\frac{n}{SP}* \frac{\left(m-ae\right)^2+n^{2}-m^{2}-n^{2}}{2*S'P*ae} + \frac{n}{S'P}* \frac{\left(m+ae\right)^2+n^{2}-m^{2}-n^{2}}{2*SP*ae} \) Area = \(\frac{1}{2} * SP * S'P * \left(\frac{n}{SP} * \frac{\left(m-ae\right)^2+n^{2}-m^{2}-n^{2}}{2*S'P*ae} + \frac{n}{S'P}*\frac{\left(m+ae\right)^2+n^{2}-m^{2}-n^{2}}{2*SP*ae}\right)\) Simplify the equation: Area = \(\frac{n}{4ae}*(\left(m-ae\right)^2+\left(m+ae\right)^2-2*(m^2+n^2))\)

Simplify the above equation and compare it with the given options: Area = \(\frac{n}{4ae}*(2a^2e^2)\) Area = be\(\sqrt{a^2 - m^2}\) Hence, the correct option is (d): be \(\sqrt{a^2 - m^2}\).