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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2020 · Mathematics · Application of Derivatives

The equation of the tangent to the curve y=b e^ -x/a at the point...

The equation of the tangent to the curve y=be−x/ay=b e^{-x/a} at the point where it crosses the yy -axis is

Options

  1. Axa−yb=1\frac{x}{a}-\frac{y}{b}=1
  2. Bax+by=1a x+b y=1
  3. Cax−by=1a x-b y=1
  4. Dxa+yb=1\frac{x}{a}+\frac{y}{b}=1

Correct answer

D. xa+yb=1\frac{x}{a}+\frac{y}{b}=1

Step-by-step solution

y=be−x/ay=b e^{-x / a} meets the yy -axis at (0,b)(0, b). Again, dydx=be−x/a\frac{d y}{d x}=b e^{-x / a}\left(-\frac{1}{a}\right)\operatorname{At}(0, b), \frac{d y}{d x}=b e^{0}\left(-\frac{1}{a}\right)=-\frac{b}{a}Therefore,requiredtangentisTherefore, required tangent isy-b=-\frac{b}{a}(x-0) \text{ or } \frac{x}{a}+\frac{y}{b}=1
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