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

Answers (1)


                \pm 10


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


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

To find:

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


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

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