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  • Let <img alt="\mathrm{x \tan \alpha+y \sin \alpha=\alpha}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%20%5Ctan%20%5Calpha&plus;y%20%5Csin%20%5Calpha%3D%5Calpha%7D"

Let \mathrm{x \tan \alpha+y \sin \alpha=\alpha} and  \mathrm{a x \operatorname{cosec} a+y \cos \alpha=1} be two variable straight lines, \mathrm{\alpha } being the parameter, Let P be the pouint of intersection of the lines. In the limitting position when \mathrm{a \rightarrow 0 } , the point is 

Option: 1

P(2,1)


Option: 2

P(-2,1)


Option: 3

(-2,-1)


Option: 4

(2,-1)


Answers (1)

best_answer

The given equation can be written as 

\mathrm{x \sin \alpha+y \sin \alpha \cos a=\alpha \cos a} --------(i)

and \mathrm{a x+y \sin a \cos a-\sin a}-------------(ii)

substracting Eq. (ii) from (i) , then 

\mathrm{\begin{aligned} \therefore \quad & =\frac{\alpha \cos \alpha-\sin \alpha}{\sin \alpha-\alpha} \\ & =\lim _{a \rightarrow 0} \frac{\alpha \cos \alpha-\sin \alpha}{\sin \alpha-\alpha} \\ & \frac{-\alpha(1-\cos \alpha)}{(\sin \alpha-\alpha)}+\frac{(\alpha-\sin \alpha)}{(\sin \alpha-\alpha)} \\ & =\lim _{a \rightarrow 0} \frac{-\frac{(1-\cos \alpha)}{\alpha^2}}{\left(\frac{\sin \alpha-\alpha}{\alpha^3}\right)}-1 \\ & =\frac{-1 / 2}{-1 / 6}-1 \\ \end{aligned}}

            \mathrm{\begin{aligned} & =3-1 \\ & =2 \end{aligned}}

From eq. \mathrm{2 \sin \alpha+y \sin \alpha \cos \alpha=\alpha \cos \alpha}

\mathrm{\begin{aligned} y & =\lim _{a \rightarrow 0} \frac{\alpha \cos \alpha-2 \sin \alpha}{\sin \alpha \cos \alpha} \\ & =\lim _{a \rightarrow 0} \frac{\frac{\alpha}{\sin \alpha} \cdot \cos \alpha-2}{\cos \alpha} \end{aligned}}

\mathrm{\begin{aligned} & =\frac{1 \times \cos 0-2}{1}=\frac{1-2}{1}=-1 \\ P & =(2,-1) \end{aligned}}

 

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shivangi.shekhar

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