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ઉપવલય $4x^2 + 9y^2 = 1$ ઉપર કયા બિંદુ આગળના સ્પર્શકો $8x = 9y$ ને સમાંતર હોય ?
$\left( {\frac{2}{5},\,\,\frac{1}{5}} \right)$
$\left( { - \frac{2}{5},\,\,\frac{1}{5}} \right)$ અથવા $\,\left( {\frac{2}{5},\,\, - \frac{1}{5}} \right)\,\,$
$\left( { - \frac{2}{5},\,\,\frac{{ - 1}}{5}} \right)$
$\left( {\frac{{ - 3}}{5},\,\,\frac{{ - 2}}{5}} \right)$
Solution
ઉપવલય $\frac{{{x^2}}}{{1/4}}\,\, + \;\,\frac{{{y^2}}}{{1/9}}\,\, = \,\,1\,\,$
$ \Rightarrow \,\,{a^2}\,\, = \,\,\frac{1}{4},\,\,{b^2}\,\, = \,\,\frac{1}{9}$
તેના સ્પર્શકનું સમીકરણ $4xx'\,\, + \,\,\,9yy'\,\, = \,\,1\,\,$
$\therefore \,\,m\,\, = \,\, – \,\,\frac{{4x'}}{{9y'}}\,\, = \,\,\frac{8}{9}\,\,$
$ \Rightarrow \,\,x'\,\, = \,\, – 2y'\,$ અને
$4x{'^2}\,\, + \;\,9y{'^2}\,\, = \,\,1\,\, \Rightarrow \,\,4\,\,{\left( {x'} \right)^2}\,\, + \;\,9\,\,\frac{{{{\left( {x'} \right)}^2}}}{4}\,\, = \,\,1\,\, \Rightarrow \,\,x'\,\, = \,\, \pm \,\,\frac{2}{5}$
જ્યારે $x'\,\, = \,\,\frac{2}{5},\,\,y' = \,\, – \frac{1}{5}\,$ અને જ્યારે $\,x'\,\, = \,\, – \frac{2}{5},\,\,y'\,\, = \,\,\frac{1}{5}$
આથી બિંદુઓ $\left( {\frac{2}{5},\,\, – \frac{1}{5}} \right)\,\,$ અને $ \left( { – \frac{2}{5},\,\,\frac{1}{5}} \right)\,\,$ હોય.
