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Q.
If the hyperbola $x y=8$ and ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ have a common tangent, then
Conic Sections
Solution:
Equation of tangent to the ellipse having slope ' $m$ ' is
$y=m x \pm \sqrt{16 m^{2}+b^{2}} $
Solving it with the hyperbola $x y=8$,
$x\left(m x \pm \sqrt{16 m^{2}+b^{2}}\right)=8 $
or $m x^{2} \pm \sqrt{16 m^{2}+b^{2}} x-8=0$
Since line is tangent to the hyperbola, above equation must have equal roots.
$\therefore D=0 $
$\therefore 16 m^{2}+b^{2}+32 m \geq 0, \forall m \in R $
$\therefore (32)^{2}-4(16)\left(b^{2}\right) \leq 0 $
$\therefore b^{2} \geq 16 $
$\therefore |b| \geq 4$