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ઉપવલય $x^{2} + 2y^{2} = 2$ ના કોઈ પણ સ્પર્શકનો અક્ષો વચ્ચે કપાયેલ અંત:ખંડના મધ્યબિંદુનો બિંદુપથ મેળવો.
$\frac{1}{{2{x^2}}}\,\, + \,\frac{1}{{4{y^2}}}\,\, = \,\,1$
$\frac{1}{{4{x^2}}}\,\, + \,\,\frac{1}{{2{y^2}}}\,\, = \,\,1$
$\frac{{{x^2}}}{2}\,\, + \;\,\frac{{{y^2}}}{4}\,\, = \,\,1$
$\frac{{{x^2}}}{4}\,\, + \;\,\frac{{{y^2}}}{2}\,\, = \,\,1$
Solution
ધરોકે સ્પર્શ બિંદુ ${\text{R}}\,\, \equiv \,\,\left( {\sqrt {\text{2}} \,\,\cos \theta ,\,\,\sin \,\,\theta } \right)\,$
સ્પર્શક $AB$ નું સમીકરણ $\frac{x}{{\sqrt 2 }}\,\,\cos \,\,\theta \,\, + \;\,y\sin \,\,\theta \,\, = \,\,1$
$A\,\, \equiv \,\,\left( {\sqrt 2 \,\,\sec \,\theta ,\,\,0} \right);\,\,B\,\, \equiv \,\,\left( {0,\,\,\cos ec\,\,\theta } \right)$
ધરોકે $AB\,\,$ નું મધ્યબિંદુ $Q\,\,\left( {h,\,\,k} \right)$ છે
$ \Rightarrow \,\,h\,\, = \,\,\frac{{\sec \,\,\theta }}{{\sqrt 2 }},\,\,k\,\, = \,\,\frac{{\cos ec\,\,\theta }}{2}$
$ \Rightarrow \,\,\cos \,\,\theta \,\, = \,\,\frac{1}{{h\,\sqrt 2 }}$ ; $\,\,\sin \,\,\theta \,\, = \,\,\frac{1}{{2k}}\, $
$\Rightarrow \,\,\frac{1}{{2{h^2}}}\,\, + \,\,\frac{1}{{4{k^2}}}\,\, = \,\,1,\,\,$
મંગેલો બિંદુપથ $\frac{1}{{2{x^2}}}\,\, + \;\,\frac{1}{{4{y^2}}}\,\, = \,\,1$
