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(B) The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. The equation of a tangent to the ellipse with slope $m$ is given by $y=mx \pm

Vedclass · Accepted Answer

$9$

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(B) The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. The equation of a tangent to the ellipse with slope $m$ is given by $y=mx \pm \sqrt{a^{2}m^{2}+b^{2}}$.Substituting $a^{2}=16$ and $b^{2}=9$,we get $y=mx \pm \sqrt{16m^{2}+9}$ $(i)$.The equation of the circle is $x^{2}+y^{2}=12$. The equation of a tangent to the circle with slope $m$ is given by $y=mx \pm r\sqrt{1+m^{2}}$.Substituting $r^{2}=12$,we get $y=mx \pm \sqrt{12(1+m^{2})}$ $(ii)$.For the common tangent,the constant terms in $(i)$ and $(ii)$ must be equal:$16m^{2}+9 = 12(1+m^{2})$$16m^{2}+9 = 12+12m^{2}$$16m^{2}-12m^{2} = 12-9$$4m^{2} = 3$Multiplying both sides by $3$,we get $12m^{2} = 9$.

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

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