Question

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

• A. 6
• B. 9
• C. 10
• D. 12

Answer 1

Answer: (B)

$\frac{x^2}{16}+\frac{y^2}{9}=1$
equation of tangent to the ellipse is
$y=mx\pm \sqrt{a^2{m}^2+b^2}$
$y=mx\pm \sqrt{16m^2+9}$ ...(i)
$x^2+y^2=12$
equation of tangent to the circle is
$y=mx\pm \sqrt{12}\sqrt{1+m^2}$ ...(ii)
for common tangent equate eq. (i) and (ii)
$\Rightarrow 16m^2+9=12\left(1+m^2\right)$
$16m^2-12m^2=3$
$4m^2=3$
$12m^2=9$