Question

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

• A. $5x+y=2$
• B. $5x-y=2$
• C. $x–5y=2$
• D. $x+5y=2$

Answer 1

The correct option is D $x+5y=2$

Given : Equation of the curve
$y(1+x^2)=2-x$
It is given that the curve crosses $x-$ axis
Substituting $y=0$ in equation (1),

$\therefore 0(1+x^2)=2-x$

$\Rightarrow x=2$

So, curve is passing through the point $(2,0)$

Differentiating eq (1) both sides w.r.t $x$

$\Rightarrow y(0+2x)+(1+x^2)\frac{dy}{dx}=0-1$
$\Rightarrow \frac{dy}{dx}=\frac{-1-2xy}{1+x^2}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,0)}=\frac{-1-2\times 0}{1+2^2}=-\frac{1}{5}$

Slope of tangent to the curve $=-\frac{1}{5}$

Equation of straight line passing through $(2,0)$ is

$y-0=-\frac{1}{5}(x-2)$

$[\because \text{Eq. of tangent}:y-y_1=m(x-x_1)]$

$\Rightarrow 5y+x=2$

The equation of tangent is : $5y+x=2$
Hence, option $a$ is correct.