#### Provide Solution for RD Sharma Class 12 Chapter 15 Tangents and Normals  Exercise Fill in the blanks  Question 5

$\left ( 6,7 \right )$

Hint:

First we find slope of curve then equate with given slope  $\frac{2}{5}$

Given:

Given curve, $y= 2+\sqrt{4x+1}$  and tangent has slope  $\frac{2}{5}$

To find:

We have to find the coordinates of the point on the given curve where tangent has slope $\frac{2}{5}$ .

Solution:

We have,

$y= 2+\sqrt{4x+1}$                                                                                                                                  … (i)

Differentiating equation (i) on both side with respect to $x$ , we get

\begin{aligned} &\Rightarrow \quad y=2+(4 x+1)^{\frac{1}{2}} \\\\ &\Rightarrow \quad \frac{d y}{d x}=0+\frac{1}{2}(4 x+1)^{\frac{1}{2}-1} \frac{d(4 x+1)}{d x} \end{aligned}                                                                         $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\frac{1}{2}(4 x+1)^{\frac{-1}{2}}(4) \\\\ &\Rightarrow \quad \frac{d y}{d x}=\frac{2}{\sqrt{4 x+1}} \end{aligned}

And tangent has slope  $\frac{2}{5}$

\begin{aligned} &\Rightarrow \quad \frac{2}{\sqrt{4 x+1}}=\frac{2}{5} \\\\ &\Rightarrow \quad \frac{10}{2}=\sqrt{4 x+1} \end{aligned}

Squaring on both sides,

$\begin{array}{ll} \Rightarrow & (5)^{2}=4 x+1 \\\\ \Rightarrow & 25=4 x+1 \\\\ \Rightarrow & x=6 \end{array}$

Substituting the value of  $x= 6$  in equation (i), we get

$\Rightarrow \quad y=2+\sqrt{4(6)+1}$

$\begin{array}{ll} \Rightarrow & y=2+\sqrt{25} \\\\ \Rightarrow & y=2+5=7 \end{array}$

Hence the required coordinate is  $\left ( 6,7 \right )$