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ધારો કે $S$ અને $S'$ નાભિઓ વાળા ઉપવલય $\frac{{{x^2}}}{{25}}\,\, + \;\,\frac{{{y^2}}}{{16}}\,\, = \,\,1$પરંતુ ચલ બિંદુ $P$ છે. જો ત્રિકોણ $PSS'$ નું ક્ષેત્રફળ $A$ નું મહત્તમ મૂલ્ય : ............. ચો. એકમ
$12$
$24$
$36$
$48$
Solution
Equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$
$e=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{25-16}}{5}=\frac{\sqrt{9}}{5}=\frac{3}{5}$
$\Rightarrow a e=5 \times \frac{3}{5}=3$
Foci $F_1$ and $F_2$ are $(-3,0)$ and $(3,0)$ respectively.
$\text { Area }=\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\-3 & 0 & 1 \\3 & 0 & 1 \end{array}\right|$ $=\frac{1}{2}[ x (0)- y (-3-3)+1(0)]$
$=\frac{-6 y }{2}=-3 y =3 \times 4 \times \sqrt{1-\frac{ x ^2}{25}} \text { since } y =4 \sqrt{1-\frac{ x ^2}{25}}$
$A=12 \times \sqrt{1-\frac{x^2}{25}}$
for maximum value of $A$
$\frac{ d ( A )}{ dx }=0$
$\Rightarrow \frac{ d ( A )}{ dx }=12 \times \frac{1}{2 \sqrt{1-\frac{ x ^2}{25}}} \times \frac{-2 x }{25}=0$
$\Rightarrow x =0$
So $A$ is $\max$ when $x =0$
$\Rightarrow A _{\max }=3 \times 4 \times \sqrt{1-0}=4 \times 3 \times \sqrt{1}=4 \times 3=12 \text { sq.units }$
$\therefore$ the maximum area is $12$ sq.units
