જો P (x, y), F_1 = (3, 0), F_2 (-3, 0) અને 1 | vedclass

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Question

વક્રનું સમીકરણ $\,\,\frac{{{x^2}}}{{{5^2}}}\,\,\, + \;\,\frac{{{y^2}}}{{{4^2}}}\,\, = \,\,1$

$ \Rightarrow \,\, - 5\,\, \leqslant \,\,x\,\, \l

Vedclass · Accepted Answer

$10$

Vedclass · Answer

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,\,$

$\,\left( {\because \,\,25\,\, - \,\,3x\,\, > \,\,0,\,\,25\,\, + \;\,3x\,\, > \,\,0} \right)$

જો $P (x, y), F_1 = (3, 0), F_2 (-3, 0) $ અને $16x^{2} + 25y^{2} = 400$ તો $PF_1 + PF_2 = …....$

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