Question

The equation of the normal to the curve $y(1+x^2)=2-x$ where the tangent crosses $x$-axis is

• A. $5x-y-10=0$
• B. $x-5y-10=0$
• C. $5x+y+10=0$
• D. $x+5y+10=0$

Answer 1

The correct option is A $5x-y-10=0$

$y(1+x^2)=2-x$
$\Rightarrow y=\frac{2-x}{1+x^2}$
$\Rightarrow \frac{dy}{dx}=\frac{x^2-4x-1}{(1+x^2{)}^2}$
The point where the tangent crosses $x$-axis is $(2,0).$
Slope of the tangent at $(2,0)$ is $\frac{dy}{dx}{]}_{(2,0)}=\frac{-1}{5}$
Therefore, the equation of the normal at $(2,0)$ is
$y-0=5(x-2)$
$\Rightarrow 5x-y-10=0$