"If m is the slope of a common tangent to the curves ${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${x^2} + {y^2} = 12$, then $12{m^2}$ is equal to :"

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Accepted Answer

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${C_1}:{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${C_2}:{x^2} + {y^2} = 12$

Let $y = mx \pm \,\sqrt {16{m^2} + 9}$ be any tangent to C₁ and if this is also tangent to C₂ then

$$\left| {{{\sqrt {16{m^2} + 9} } \over {\sqrt {{m^2} + 1} }}} \right| = \sqrt {12} $$

$\Rightarrow 16{m^2} + 9 = 12{m^2} + 12$

$\Rightarrow 4{m^2} = 3 \Rightarrow 12{m^2} = 9$

"

If m is the slope of a common tangent to the curves ${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${x^2} + {y^2} = 12$, then $12{m^2}$ is equal to :

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If m is the slope of a common tangent to the curves ${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${x^2} + {y^2} = 12$, then $12{m^2}$ is equal to :

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