Question

The equation of the line parallel to x-axis and tangent to the curve $y=\frac{1}{x^2+2x+5}$ is

• A. $y=\frac{1}{4}$
• B. $y=4$
• C. $y=\frac{1}{2}$
• D. $y=0$
• E. $y=2$

Answer 1

Answer: (A)

Curve, $y=\frac{1}{x^2+2x+5}$ ..(i)
Let the equation of line which is parallel to $x-$ axis is, $y=c$ ...(ii) The line (ii) is a tangent to curve (i), then slope of curve = slope of line
$\frac{-(2x+2)}{{(x^2+2x+5)}^2}=0$ $\left(\because \frac{dy}{dx}=\frac{-(2x+2)}{{(x^2+2x+5)}^2}\right)$
$\Rightarrow$ $x=-1$ From Eq. (i),
$y=\frac{1}{1-2+5}=\frac{1}{4}$ From Eq. (ii), $c=\frac{1}{4}$
Hence, the required equation of line is, $y=\frac{1}{4}.$