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 $5x-y-10=0$

Given curve can be written as, $y=\frac{2-x}{1+x^2}$

To get the point where the above curve will cross the x-axis, put $y=0$

$\Rightarrow 0=\frac{2-x}{1+x^2}\Rightarrow x=2$

Thus the required point is $(2,0)$

Now differentiate the given curve using quotient rule,

$\frac{dy}{dx}=\frac{(1+x^2)(-1)-(2-x)\left(2x\right)}{(1+x^2{)}^2}$

$=\frac{x^2-4x-1}{(1+x^2{)}^2}$

Thus the slope of a tangent to this curve at point $(2,0)$ is

$m=\frac{dy}{dx}{|}_{(2,0)}=\frac{2^2-4\left(2\right)-1}{(1+2^2{)}^2}=-\frac{5}{25}=-\frac{1}{5}$

Therefore the slope of normal at this point will be $5$

Hence required equation of normal is given by, $(y-0)=5(x-2)$

$\Rightarrow 5x-y-10=0$