Question

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

• A. $x+5y=2$
• B. $x-5y=2$
• C. $5x–y=2$
• D. $5x+y–2=0$

Answer 1

The correct option is A

$x+5y=2$

Explanation for the correct answer:

Step 1: Find the first order derivative of the equation of the given curve

Given, Equation of curve $(1+x^2)y=2–x....\left(\mathrm{i}\right)$

On differentiating eq.$\left(\mathrm{i}\right)$ with respect to $x$ we get,

$$\begin{gathered} (1+x^2)\frac{dy}{dx}+y\left(0+2x\right)=-1 \\ \Rightarrow (1+x^2)\frac{dy}{dx}+2xy=-1 \\ \Rightarrow \frac{dy}{dx}=\frac{-1-2xy}{1+x^2} \end{gathered}$$

Step 2:Find the slope of the tangent at the required point

Since, the given curve passes through $x$− axis, that is, $y=0$

By substituting the value of $y$ in equation $\left(\mathrm{i}\right)$ we get,

$$\begin{gathered} 0(1+x^2)=2-x \\ \Rightarrow x=2 \end{gathered}$$

So, the curve passes through the point $(2,0)$.

The first order derivative at a given point gives the slope of the tangent at that point

$\therefore$Slope of tangent at $(2,0)$ is ${\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}=m=\frac{-1\times -2\times 0}{1+2^2}$

$m=\frac{-1}{5}$

Step 3: Find the equation of the tangent using slope-point form

Therefore, equation of tangent of the curve passing through $(2,0)$ is

$$\begin{gathered} y-0=\frac{-1}{5}(x-2) \\ \Rightarrow 5y=-x+2 \\ \Rightarrow x+5y=2 \end{gathered}$$

Hence, the correct answer is option (A).