A) \[{{x}^{2}}+2xy\sec 2\alpha +{{y}^{2}}=0\]
B) \[{{x}^{2}}+2xy\,\text{cosec}\,2\alpha +{{y}^{2}}=0\]
C) \[{{x}^{2}}-2xy\,\text{cosec}\,2\alpha +{{y}^{2}}=0\]
D) \[{{x}^{2}}-2xy\sec 2\alpha +{{y}^{2}}=0\]
Correct Answer: D
Solution :
Any line through the origin is\[y=mx\]. If it makes an angle \[\alpha \]with the line\[y=x\], then we should have \[\tan \alpha =\pm \left\{ \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right\}=\pm \frac{(m-1)}{1+m}\] or \[{{(1+m)}^{2}}{{\tan }^{2}}\alpha ={{(m-1)}^{2}}\] \[\Rightarrow {{m}^{2}}-2m\left\{ \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha } \right\}+1=0\] \[\Rightarrow {{m}^{2}}-2m\sec 2\alpha +1=0\], \[\left\{ \because \frac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }=\sec 2\alpha \right\}\] But \[m=\frac{y}{x},\]hence on eliminating m, we get the required equation \[{{y}^{2}}-2xy\sec 2\alpha +{{x}^{2}}=0\]
You need to login to perform this action.
You will be redirected in
3 sec
