A) \[\frac{\sqrt{3}}{2}\]
B) \[\frac{1}{2}\]
C) \[-\frac{\sqrt{3}}{2}\]
D) \[\frac{3}{8}\]
Correct Answer: D
Solution :
We know that the equation of the normal at point \[(a\cos \theta ,\,b\sin \theta )\] on the curve \[{{x}^{2}}+\frac{{{y}^{2}}}{4}=1\] is given by \[ax\sin \theta -by\text{cosec }\theta ={{a}^{2}}-{{b}^{2}}\] .....(i) Comparing equation (i) with \[2x-\frac{8}{3}\lambda y=-3\]. We get, \[a\sin \theta =2\], \[b\text{ cosec}\theta =\frac{8}{3}\lambda \] or \[ab=\frac{16}{3}\lambda \] .....(ii) \[\because \,a=1,\,b=2\]; \\[2=\frac{16}{3}\lambda \] or \[\lambda =3/8\]
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