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  • Let   be a common tangent line to the curves  <img alt="\mathrm{4 x^2+9 y^2=36}" src="h

Let  \mathrm{L} be a common tangent line to the curves  \mathrm{4 x^2+9 y^2=36}  and   \mathrm{(2 x)^2+(2 y)^2=31} . Then the square of the slope of the line  is \mathrm{L} 

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

Given curves are  \mathrm {\frac{x^2}{9}+\frac{y^2}{4}=1\: and \: x^2+y^2=\frac{31}{4}} 
Let slope of common tangent be  \mathrm {m.}
So, tangents of given ellipse and circle are respectively
\mathrm { y=m x \pm \sqrt{9 m^2+4} \text { and } y=m x \pm \frac{\sqrt{31}}{2} \sqrt{1+m^2} .} 
Hence, \mathrm { 9 m^2+4=\frac{31}{4}\left(1+m^2\right) \\ \Rightarrow 36 m^2+16=31+31 m^2 \Rightarrow 5 m^2=15 \Rightarrow m^2=3 .}

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Nehul

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