10.2 parabola,ellipse,hyperbola — ગણિત ધોરણ 11 સાયન્સ — Question

$\,\frac{{{x^2}}}{{16}}\,\, + \;\,\frac{{{y^2}}}{9}\,\, = \,\,1$

આને  $\,\frac{{{x^2}}}{{{a^2}}}\,\, + \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,\,$ સાથે સરખાવતા 

${a^2}\,\, = \,\,16$ અને  ${b^2}\,\, = \,\,9\,$ મળે

અને રેખા $y\,\, = \,\,x\,\, + \;\,\lambda \,$ ને  $\,y = \,\,mx\,\, + \;\,c\, $ 

સાથે સરખાવતા  $\therefore \,\,m\,\, = \,\,1\,$  અને  $c\,\, = \,\,\lambda $

જો રેખા   $y\,\, = \,\,x\,\, + \;\,\lambda $ ઉપવલય $\,9{x^2}\,\, + \;\,16{y^2}\,\, = \,\,144\,\,$ ને સ્પર્શે , તો $\,{c^2}\,\, = \,\,{a^2}\,{m^2}\,\, + \;\,{b^2}$

$ \Rightarrow \,\,{\lambda ^2}\,\, = \,\,16\,\, \times \,\,{1^2}\,\, + \;\,9\,\, \Rightarrow \,\,{\lambda ^2}\,\, = \,\,25\,\,\,\,\therefore \,\,\lambda \,\, = \,\, \pm \,\,5$