Let `y=f(x)` be drawn with `f(0) =2` and for each real number `a` the line tangent to `y = f(x)` at `(a,f(a))` has x-intercept ` (a-2)`. If `f(x)` is of the form of `k e^(px)` then`k/p` has the value equal to

Answer

To solve the problem step by step, we will follow the given information and derive the required values.

### Step 1: Identify the function form
Given that \( f(x) = k e^{px} \).

### Step 2: Use the initial condition
We know that \( f(0) = 2 \). Substituting \( x = 0 \) into the function:
\[ f(0) = k e^{p \cdot 0} = k e^0 = k \]
Thus, we have:
\[ k = 2 \]
So, the function simplifies to:
\[ f(x) = 2 e^{px} \]

### Step 3: Find the derivative of the function
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2 e^{px}) = 2p e^{px} \]

### Step 4: Determine the tangent line at point \( (a, f(a)) \)
The slope of the tangent line at point \( (a, f(a)) \) is given by:
\[ f'(a) = 2p e^{pa} \]
The equation of the tangent line at this point can be expressed as:
\[ y - f(a) = f'(a)(x - a) \]
Substituting \( f(a) = 2 e^{pa} \) and \( f'(a) = 2p e^{pa} \):
\[ y - 2 e^{pa} = 2p e^{pa}(x - a) \]

### Step 5: Find the x-intercept of the tangent line
To find the x-intercept, set \( y = 0 \):
\[ 0 - 2 e^{pa} = 2p e^{pa}(x - a) \]
This simplifies to:
\[ -2 e^{pa} = 2p e^{pa}(x - a) \]
Dividing both sides by \( 2 e^{pa} \) (assuming \( e^{pa} \neq 0 \)):
\[ -1 = p(x - a) \]
Thus, we can express \( x \) in terms of \( a \):
\[ x - a = -\frac{1}{p} \implies x = a - \frac{1}{p} \]

### Step 6: Relate the x-intercept to the given condition
According to the problem, the x-intercept is given as \( a - 2 \). Therefore, we have:
\[ a - \frac{1}{p} = a - 2 \]
This leads to:
\[ -\frac{1}{p} = -2 \implies \frac{1}{p} = 2 \implies p = \frac{1}{2} \]

### Step 7: Calculate \( k/p \)
We already found \( k = 2 \) and \( p = \frac{1}{2} \). Now, we can calculate:
\[ \frac{k}{p} = \frac{2}{\frac{1}{2}} = 2 \times 2 = 4 \]

### Final Answer
The value of \( \frac{k}{p} \) is:
\[ \boxed{4} \]