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Question

The slopes of the common tangents of the ellipse $$\displaystyle \frac{x^2}{4} + \frac{y^2}{1} = 1$$ and the circle $$x^2 + y^2 = 3$$ are


A
±1
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B
±2
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C
±3
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D
None of the above
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Solution

The correct option is D $$\pm \sqrt{2}$$
Let $$m$$ be the slope of the common tangent. 
For the ellipse, the equation of the tangent is $$ y = mx \pm \sqrt{4m^2 + 1} $$
For the circle, the equation of the tangent can be written as $$ y= mx \pm \sqrt{3(1+m^2)} $$
As both the equations represent the same tangent, 
$$\pm \sqrt{3} \sqrt{1 + m^2} = \pm \sqrt{4m^2 + 1}$$
$$\Rightarrow   3 + 3m^2 = 4m^2 + 1$$
$$\Rightarrow  m^2 = 2$$
$$\Rightarrow  m = \pm \sqrt{2}$$

Mathematics

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