Question

The area of the triangle formed by the tangent to the curve $y=\frac{8}{4+x^2}$ at $x=2$ on it and the $x$-axis is

• A. $2$ sq.units
• B. $4$ sq.units
• C. $8$ sq.units
• D. $16$ sq.units

Answer 1

Answer: (B)

$4$ sq.units

$y^{\prime }=\frac{8\times 24}{(4+x^2{)}^2}$
$y=1=\frac{-8\times 2\times 2}{(4+4{)}^2}$
$=\frac{-5\times 4}{64}=\frac{1}{2}$
$(y-1)=\frac{-1}{2}(x-2)$
$2y-2=-x+2$
$2y+x=4$
$A=\frac{1}{2}\times 2\times 4=4$