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

$\frac{-b}{a}$

Hint:

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:

Given curve,

$y=b e^{\frac{-x}{a}}$

To find:

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

Solution:

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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