If m is the slope of a common tangent to the curves x ^ 2 16 + y ^...

Given,

$\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{9} = 1$

we know that equation of tangent to the ellipse is

$y = mx \pm \sqrt{{a}^{2}{m}^{2} + {b}^{2}}$

$y = mx \pm \sqrt{16{m}^{2} + 9} . . . \left(1\right)</p><p>${x}^{2} + {y}^{2} = 12$</p><p>Also equation of tangent to the circle is</p><p>$y = mx \pm \sqrt{12}\sqrt{1 + {m}^{2}} . . . \left(2\right)$</p><p>Now, for common tangent equate eq.\left(1\right) and \left(2\right)</p><p>$⇒ 16{m}^{2} + 9 = 12\left(1 + {m}^{2}\right)$</p><p>$16{m}^{2} - 12{m}^{2} = 3$</p><p>$4{m}^{2} = 3$</p><p>$12{m}^{2} = 9$