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Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 8 Maths Textbook Solution.

Answers (1)




Use  \left ( 0,y_{1} \right ) is the point where tangent to the curve crosses  y- axis, then proceed next to find slope of tangent.


Given curve,

                y=b e^{\frac{-x}{a}}

To find:

We have to find the slope of tangent where the curves crosses  y- axis.


Given equation of curve is,

            y=b e^{\frac{-x}{a}}                                                                                                                  … (i)

Let  p\left(0, y_{1}\right) be the point where tangent to curve crosses  y- axis.

\Rightarrow \quad y_{1}=b

So the point p is p\left ( 0,b \right )

Differentiating equation (i), we get

\Rightarrow \quad \frac{d y}{d x}=b e^{\frac{-x}{a}}=\frac{be}{a}^{-\frac{x}{a}}=\frac{-y}{a}      

Hence slope of tangent at  p\left ( 0,b \right ) is  \frac{-b}{a}.

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