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
B
±2
C
±3
D
None of the above

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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