#### Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 14 Maths Textbook Solution.

Equation of tangent is  $y=3$

Hint:

Tangent is parallel to X-axis so slope becomes 0

Given:

Given curve   $y=x+\frac{4}{x^{2}}$

To find:

We have to find the equation of the tangent to the given curve that is parallel to X-axis

Solution:

Given tangent is parallel to X-axis so slope is 0

$\Rightarrow \frac{d y}{d x}=0$                                                                                                                                          … (i)

Here   $y=x+\frac{4}{x^{2}}$                                                                                                                             … (ii)

On differentiate both side with respect to  $x$  we get

$\frac{d y}{d x}=1-\frac{8}{x^{3}}$

Substituting the value of  $\frac{dy}{dx}$  in equation (i), we get

\begin{aligned} &1-\frac{8}{x^{3}}=0 \\\\ &x^{3}=8 \\\\ &x=2 \end{aligned}

$\because$ From (ii) we get  $y=2+\frac{4}{4}=3$

Hence equation of tangent is $y= 3$                                         [ $\because$ equation of tangent $y-3=0(x-2)$ ]