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}. $