Question

The equation of the tangent to the curve \[y = 4{e^{ - \dfrac{x}{4}}}\] at the point where the curve crosses \[Y\] axis is equal to
A) \[3x + 4y = 16\]
B) \[4x + y = 4\]
C) \[x + y = 4\]
D) \[3x + 4y = 25\]

Answer 1

Hint:
Here we need to find the equation of the tangent to the given curve. We will first find the point at which the given curve crosses the \[Y\] axis. Then we will find the slope of the tangent at that point by differentiating the given curve with respect to the required variable. Then we will find the equation of the tangent using the obtained point and the obtained slope of the tangent.

Complete Step by step Solution:
Here we need to find the equation of the tangent to the given curve i.e. \[y = 4{e^{ - \dfrac{x}{4}}}\]
We will first find the point at which the given curve crosses the \[Y\] axis as it is given that the given curve crosses \[Y\] axis. So the value of the \[x\] coordinate will be equal to zero.
We will substitute the value of the \[x\] coordinate as zero in the equation of the curve.
\[ \Rightarrow y = 4{e^{ - \dfrac{0}{4}}} = 4{e^0}\]
On further simplification, we get
\[ \Rightarrow y = 4 \times 1 = 4\]
Therefore, the point on the \[Y\] axis at which the given curve cuts it will be equal to \[\left( {0,4} \right)\]
Now, we will find the slope of the required tangent to the curve. For that, we will differentiate the given curve with respect to \[x\] .
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {4{e^{ - \dfrac{x}{4}}}} \right)\]
On differentiating the variable terms, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 4 \times \dfrac{{ - 1}}{4} \times {e^{ - \dfrac{x}{4}}}\]
On multiplying the terms, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - {e^{ - \dfrac{x}{4}}}\]
Now, we will find the value of the slope of the tangent at the point \[\left( {0,4} \right)\]. For that, we will substitute the value of coordinate of point here.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - {e^{ - \dfrac{0}{4}}} = - 1 \times 1 = - 1\]
Thus, the slope of the tangent to the curve is equal to -1.
Now, we will find the equation of the required tangent to the curve.
We know that the general equation of a line passing through any point say \[\left( {{x_1},{y_1}} \right)\] is given by
\[ \Rightarrow y - {y_1} = {\left. {\dfrac{{dy}}{{dx}}} \right|_{\left( {{x_1},{y_1}} \right)}} \times \left( {x - {x_1}} \right)\]
Now, we will substitute the value of the slope of the tangent and the coordinates of the point here.
\[ \Rightarrow y - 4 = - 1 \times \left( {x - 0} \right)\]
On further simplification, we get
\[ \Rightarrow x + y = 4\]
Therefore, the equation of the tangent to the given curve is equal to \[x + y = 4\]

Hence, the correct option is option C.

Note:
To solve this problem, we need to remember the properties of the tangent and also the general equation of any line when the value of the slope and the point is known. A tangent to a curve is defined as the straight line that touches the given curve at only one point. We know that a tangent of a curve doesn’t cross the curve. The equation of tangent is different for different types of curves so we need to be careful while using the general equation.