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. $y=0$

Answer 1

Answer: (A)

$y=\frac{1}{4}$

Answer 2

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 E (i), $y=\frac{1}{1-2+5}=\frac{1}{4}$ From E (ii), $c=\frac{1}{4}$ Hence, the required equation of line is, $y=\frac{1}{4}.$