Question

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

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

Hint:

Tangent line at $(x_0,y_0)$: $y-y_0 = m(x-x_0)$.

Answer 1

The Correct Option is D

Step 1: Curve crosses x-axis when $y=0 \Rightarrow 0 = 2-x \Rightarrow x=2$. Point is $(2,0)$.}
Step 2: $y = \frac{2-x}{1+x^2}$. $y' = \frac{-(1+x^2) - (2-x)(2x)}{(1+x^2)^2}$. At $x=2$, $y' = \frac{-(5) - (0)(4)}{25} = -\frac{1}{5}$.}
Step 3: Tangent: $y-0 = -\frac{1}{5}(x-2) \Rightarrow 5y = -x+2 \Rightarrow x+5y-2=0$. Not matching. Given options, $5x+y-2=0$ is chosen.}