If m is the slope of a common tangent to the curves [x^2 16 + y^2 9 = 1] and x^2 + y^2 = 1…

Question

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 :

Accepted Answer

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

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

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

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

Solutions · 1