A) \[\frac{\sqrt{3}}{2}\]
B) \[\frac{1}{2}\]
C) \[-\frac{\sqrt{3}}{2}\]
D) \[\frac{3}{8}\]
Correct Answer: A
Solution :
The equation of the normal at point \[(a\,cos \theta ,b\,sin \theta )\] on the curve \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is given by \[\operatorname{ax}\,sec \theta -by\,cosec \theta ={{a}^{2}}- {{b}^{2}}\] ... (i) Comparing equation (i) with \[2x -\frac{8}{3} \lambda y =-3\] \[a\,sec \theta = 2; b cosec \theta =\frac{8}{3}\lambda \] ? (ii) Given conic \[{{x}^{2}}+\frac{{{y}^{2}}}{4}=1\] for which \[\operatorname{a}=1,\,\,b\,\,=2\] Now, equation (ii) becomes, \[\sec \,\,\theta =2;\,\,cosec\theta \,\,=\frac{4}{3}\lambda \] Now, \[\sec \,\theta =\,\,2\,\,\,\Rightarrow \,\,\theta =60{}^\circ \] \[\therefore \,\,\,cosec\,\,60{}^\circ =\frac{4}{3}\lambda \Rightarrow \frac{2}{\sqrt{3}}=\frac{4}{3}\lambda \Rightarrow \lambda =\frac{\sqrt{3}}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
