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Let \mathrm{2 x^2+y^2-3 x y=0} be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, then the length of \mathrm{\mathrm{OA}=n(n+\sqrt{10})}. The value of n is _________.

Option: 1

1


Option: 2

3


Option: 3

6


Option: 4

8


Answers (1)

best_answer

We have, \mathrm{2 x^2+y^2-3 x y=0}

\mathrm{ \begin{aligned} & \Rightarrow(2 x-y)(x-y)=0 \\ & \Rightarrow y=2 x, y=x \end{aligned} }

are the equations of straight lines passing through origin.

Now let the angle between tangents be \mathrm{2 \alpha.}

\mathrm{ \begin{gathered} \Rightarrow \frac{\tan 45^{\circ}+\tan 2 \alpha}{1-\tan 45^{\circ} \tan 2 \alpha}=2 \Rightarrow \frac{1+\tan 2 \alpha}{1-\tan 2 \alpha}=\frac{2}{1} \\\\ \Rightarrow \frac{2 \tan 2 \alpha}{2}=\frac{1}{3} \quad \text { (By componendo and dividendo rule) } \\\\ \Rightarrow \frac{2 \tan \alpha}{1-\tan ^2 \alpha}=\frac{1}{3} \Rightarrow \tan ^2 \alpha+6 \tan \alpha-1=0 \\\\ \therefore \quad \tan \alpha=\frac{-6 \pm \sqrt{(36+4)}}{2}=-3 \pm \sqrt{10} \\\\ =-3+\sqrt{10} \quad\left(\because 0<\alpha<\frac{\pi}{4}\right) \end{gathered} }
Now in \mathrm{ \triangle O A C, \tan \alpha=\frac{3}{O A}=\sqrt{10}-3}

 \mathrm{ \begin{aligned} & \therefore \quad O A=\frac{3}{(\sqrt{10}-3)} \frac{(\sqrt{10}+3)}{(\sqrt{10}+3)}=3(3+\sqrt{10}) \\ & \therefore n=3 \end{aligned} }

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shivangi.bhatnagar

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