#### Explain solution RD Sharma class 12 Chapter 15 Tangents and Normals exercise Fill in the blanks question 2

$\pm 10$

Hint:

Since the given curve touches $x-$ axis. i.e. $y=0$

Given:

$y=x^{2}+ax+25$                                                                                                                                           … (i)

To find:

We have to find the value of $a$  for which the given curve touches $x-$ axis.

Solution:

We know that,

Since the given curve,

$y=x^{2}+ax+25$   touches the $x-$ axis

$\Rightarrow \quad \frac{dy}{dx}=0$

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

$\Rightarrow \quad 2 x+a=0$

$\begin{array}{ll} \Rightarrow &\quad \frac{d y}{d x}=2 x+a \\ \end{array}$

$\Rightarrow$         $x=\frac{-a}{2}$

Putting the value of   $x=\frac{-a}{2}$  in equation (i), we get

Since  $y=0$

$\Rightarrow \quad \frac{a^{2}}{4}+a\left(\frac{-a}{2}\right)+25=0$

$\Rightarrow \quad \frac{a^{2}}{4}-\frac{a^{2}}{2}+25=0$

$\Rightarrow \quad a=\pm 10$

Hence, the required value of  $a$  are  $\pm 10$