Question

P is a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $S_1$ and $S_2$ . if Ais the area of the triangle $P{S}_1{S}_2$ , the maximum value of A is

• A. abe
• B. ab
• C. $\pi ab$
• D. $\frac{ab}{e}$

Answer 1

The correct option is A abe

$\begin{array}{c}letp\left(x,y\right),s\left(ae,o\right),s_2\left(-ae,o\right)bethegivenptr\\ thenareaof\Delta P{S}_1{S}_2=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\\ =\frac{1}{2}\left|x\left(0\right)+ae\left(0-y\right)+\left(-ae\right)\left(y\right)\right|\\ =\frac{1}{2}\left|-aey-aey\right|\\ =\frac{1}{2}\left|-2aey\right|\\ A=aey\\ A=ae\sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}\\ A=\frac{abe}{a}\sqrt{a^2-x^2}\\ A=be\sqrt{a^2-x^2}\\ \frac{da}{dx}=be\times \frac{1}{2\sqrt{a^2-x^2}}\times \left(-2x\right)\\ \frac{da}{dx}=-\frac{xbe}{\sqrt{a^2-x^2}}\\ \frac{d^{2}a}{d{x}^2}=-be\left[1-\sqrt{a^2-x^2}-x\times \frac{1}{2\sqrt{a^2-x^2}}\times \left(-2x\right)\right]\\ =\frac{-be\times 2a^2}{2{\left(a^2-x^2\right)}^{\frac{3}{2}}}\\ forcriticalpoint\\ \frac{da}{dx}=0\\ \Rightarrow \frac{-xbe}{\sqrt{a^2-x^2}}=0\\ \Rightarrow x=0\\ \frac{d^{2}A}{d{x}^2}atx=0=\frac{-2a^{2}be}{2a^3}<0\\ soareaofmaximumwhen\\ x=0\\ maximumarea=be\sqrt{a^2-0^2}\\ =be\times a\\ =abe\end{array}$