Q103 / 170 · Manipal (MET) 2017 · Mathematics
The equation of the tangent to the curve $\left(1+x^2\right) y=2-x$, where it crosses the $x$-axis, is
- Correct answerA$x+5 y=2$
- B$x-5 y=2$
- C$5 x-y=2$
- D$5 x+y-2=0$
Solution
The given equation of curve is $\left(1+x^2\right) y=2-x$, it meets $x$-axis at $(2,0)$ or it can be rewritten as $y=\frac{2-x}{1+x^2}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x} & =\frac{\left(1+x^2\right)(-1)-(2-x)(2 x)}{\left(1+x^2\right)^2} \\
& =\frac{-1-x^2-4 x+2 x^2}{\left(1+x^2\right)^2} \\
\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)} & =\frac{-1-4-8+8}{25}=-\frac{1}{5}
\end{aligned}$
$\therefore$ Required equation of tangent is
$(y-0)=-\frac{1}{5}(x-2) \Rightarrow x+5 y=2$
Hence, the correct answer is option: 'A'.