If β is one of the angles between the normals to the ellipse, x^{2}+3y^{2}=9 at the points     \left ( -3\sin \theta ,\; \sqrt{3}\cos \theta \right );\; \; \theta \equiv \left ( 0,\frac{\pi }{2} \right )   ; then \frac{2\cot \beta }{\sin 2\theta } is equal to : 

  • Option 1)

    \frac{2}{\sqrt{3}}

     

     

     

  • Option 2)

    \frac{1}{\sqrt{3}}

  • Option 3)

    {\sqrt{2}}

  • Option 4)

    \frac{\sqrt{3}}{4}

 

Answers (1)
H Himanshu

As we learned 

 

Angle between two lines (θ) -

\tan \Theta = \frac{m_{2}-m_{1}}{1+m_{1}m_{2}}

- wherein

Here m_{1},m_{2} are the slope of two lines

 

 \tan \beta =\left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |

for tangent 

\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-2x}{6y}=\frac{-x}{3y}

\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-3\cos \theta }{\sqrt{3}\sin \theta \cdot 3}=\frac{-\cot \theta }{\sqrt{3}}

Thus m_{1}=\sqrt{3}\tan \theta

Slope of normal  =  \frac{3y}{x}

                          = -\sqrt{3}\cot \theta

\tan \beta =\left | \frac{\sqrt{3}\left ( \tan \theta +\cot \theta \right )}{1-3} \right |=\frac{\sqrt{3}}{\sin 2\theta }

\frac{2\cot \beta }{\sin 2\theta }=\frac{2}{\sqrt{3}}


Option 1)

\frac{2}{\sqrt{3}}

 

 

 

Option 2)

\frac{1}{\sqrt{3}}

Option 3)

{\sqrt{2}}

Option 4)

\frac{\sqrt{3}}{4}

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