Question

The equation of tangent to the curve \[y=b{{e}^{-\dfrac{x}{a}}}\] at the point where it crosses the Y – axis is
(a) \[\dfrac{x}{a}-\dfrac{y}{b}=1\]
(b) \[ax+by=1\]
(c) \[ax-by=1\]
(d) \[\dfrac{x}{a}+\dfrac{y}{b}=1\]

Answer 1

Hint: We solve this problem by using the standard condition of finding the tangents of a curve.
We solve this problem by finding the coordinates of the point where the given curve crosses the Y – axis.
We use the condition that if any curve crosses the Y – axis then \[x=0\]
We use the condition that the equation of line having slope \[m\] and passes through point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given as
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]
Where \[m=\dfrac{dy}{dx}\]

Complete step by step answer:
Let us take the rough figure of the given function and the tangent at the point where it crosses Y-axis as follows

We are given that the equation of curve as
\[y=b{{e}^{-\dfrac{x}{a}}}\]
We are asked to find the equation of the tangent at which the given curve crosses the Y-axis.
We know that the condition that if any curve crosses the Y – axis then \[x=0\]
By substituting the value \[x=0\] in given curve equation we get
\[\begin{align}
  & \Rightarrow y=b{{e}^{-\dfrac{0}{a}}} \\
 & \Rightarrow y=b\times 1 \\
 & \Rightarrow y=b \\
\end{align}\]
Here, we can see that the point where the given curve touches Y – axis is \[\left( 0,b \right)\]
Now, let us take the given equation of curve that is
\[y=b{{e}^{-\dfrac{x}{a}}}\]
By differentiating the above equation with respect to \[x\] we get
\[\Rightarrow \dfrac{dy}{dx}=b\dfrac{d}{dx}\left( {{e}^{-\dfrac{x}{a}}} \right)........equation(i)\]
We know that the standard formula of derivatives that is
\[\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}\]
We also know that the chain rule of the derivatives that is
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)\]
By using the above two formula to equation (i) we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=b\left( {{e}^{-\dfrac{x}{a}}} \right)\left( \dfrac{d}{dx}\left( -\dfrac{x}{a} \right) \right) \\
 & \Rightarrow \dfrac{dy}{dx}=b\left( {{e}^{-\dfrac{x}{a}}} \right)\left( \dfrac{-1}{a} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{b}{a}{{e}^{-\dfrac{x}{a}}} \\
\end{align}\]
Now, let us substitute the point \[\left( 0,b \right)\] in above equation then we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=-\dfrac{b}{a}{{e}^{0}} \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{b}{a} \\
\end{align}\]
We know that the slope of any curve is given as \[m=\dfrac{dy}{dx}\]
By using the above condition we get the slope of tangent at \[\left( 0,b \right)\] as
\[\Rightarrow m=-\dfrac{b}{a}\]
We know that the condition that the equation of line having slope \[m\] and passes through point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given as
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]
By using the above condition to required tangent at point \[\left( 0,b \right)\] we get
\[\begin{align}
  & \Rightarrow y-b=\left( -\dfrac{b}{a} \right)\left( x-0 \right) \\
 & \Rightarrow \dfrac{y}{b}-1=-\dfrac{x}{a} \\
 & \Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=1 \\
\end{align}\]
Therefore we can conclude that the equation of tangent of given curve at point where it crosses Y – axis is given as \[\dfrac{x}{a}+\dfrac{y}{b}=1\]
So, option (d) is the correct answer.

Note:
 Students may make mistakes in considering the coordinate of point when a curve touches the Y-axis.
We have the condition that the x coordinate of a point when a curve touches Y – axis as \[x=0\]
But some students may misunderstand that it touches Y-axis so \[y=0\]
But this is not correct because touching the Y – axis means it can touch anywhere on the Y – axis. So, there will be some value for y-coordinate.