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જો $P (x, y), F_1 = (3, 0), F_2 (-3, 0) $ અને $16x^{2} + 25y^{2} = 400$ તો $PF_1 + PF_2 = …....$
$8$
$6$
$10$
$12$
Solution
વક્રનું સમીકરણ $\,\,\frac{{{x^2}}}{{{5^2}}}\,\,\, + \;\,\frac{{{y^2}}}{{{4^2}}}\,\, = \,\,1$
$ \Rightarrow \,\, – 5\,\, \leqslant \,\,x\,\, \leqslant \,\,5,\,\, – 4\,\, \leqslant \,\,y\,\, \leqslant \,\,4$
$P{F_1}\,\, + \;\,P{F_2}\,\, = \,\sqrt {\left[ {{{\left( {x\,\, – \,\,3} \right)}^2}\,\, + \,\,{y^2}} \right]} \,\, + \;\,\sqrt {\left[ {{{\left( {x\,\, + \;\,3} \right)}^2}\,\, + \;\,{y^2}} \right]} $
$ = \,\,\sqrt {{{\left( {x\,\, – \,\,3} \right)}^2}\,\, + \,\,\frac{{400\,\, – \,\,16{x^2}}}{{25}}} \,\, + \;\,\sqrt {{{\left( {x\,\, + \,\,3} \right)}^2}\,\, + \,\,\frac{{400\,\, – \,\,16{x^2}}}{{25}}} $
$ = \,\,\frac{1}{5}\,\,\left\{ {\sqrt {\left( {9{x^2}\,\, + \,625\,\, – \,\,150\,\,x} \right)} \,\,\, + \;\,\sqrt {\left( {9{x^2}\,\, + \;\,625\,\, + \;\,150\,\,x} \right)} \,} \right\}$
$ = \,\,\frac{1}{5}\,\,\left\{ {\sqrt {{{\left( {3x\,\, – \,\,25} \right)}^2}\,\, + \;\,\sqrt {{{\left( {3x\,\, + \;\,25} \right)}^2}} } } \right\}\,\,$
$ = \,\,\frac{1}{5}\,\,\left\{ {25\,\, – \,\,3x\,\, + \;\,3x\,\, + \;\,25} \right\}\,\, = \,\,10,\,$
$\,\left( {\because \,\,25\,\, – \,\,3x\,\, > \,\,0,\,\,25\,\, + \;\,3x\,\, > \,\,0} \right)$
