Question

If the hyperbola $xy=8$ and ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ have a common tangent, then

• A. $|b|\ge 4$
• B. $0<|b|\le 4$
• C. $2\le |b|\le 4$
• D. $|b|\ge 6$

Answer 1

Answer: (B)

Equation of tangent to the ellipse having slope ' $m$ ' is
$y=mx\pm \sqrt{16m^2+b^2}$
Solving it with the hyperbola $xy=8$,
$x\left(mx\pm \sqrt{16m^2+b^2}\right)=8$
or $m{x}^2\pm \sqrt{16m^2+b^2}x-8=0$
Since line is tangent to the hyperbola, above equation must have equal roots.
$\therefore D=0$
$\therefore 16m^2+b^2+32m\ge 0,\forall m\in R$
$\therefore (32{)}^2-4(16)\left(b^2\right)\le 0$
$\therefore b^2\ge 16$
$\therefore |b|\ge 4$