A normal is drawn to parabola `y^2=4x` at (1,2) and tangent is drawn to `y=e^x` at `(c,e^c)`. If tangent and normal intersect at `x-axis` then find C.

Answer

To solve the problem, we need to find the value of \( c \) such that the normal to the parabola \( y^2 = 4x \) at the point \( (1, 2) \) intersects the tangent to the curve \( y = e^x \) at the point \( (c, e^c) \) at the x-axis.

### Step 1: Find the slope of the tangent to the parabola at the point (1, 2)

The equation of the parabola is given by \( y^2 = 4x \).

To find the slope of the tangent at the point \( (1, 2) \), we can use the formula for the slope of the tangent to the parabola \( y^2 = 4ax \):

\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{2y}{4} = \frac{y}{2} \]

At the point \( (1, 2) \):

\[ \text{slope} = \frac{2}{2} = 1 \]

### Step 2: Find the slope of the normal to the parabola

The slope of the normal is the negative reciprocal of the slope of the tangent:

\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -1 \]

### Step 3: Write the equation of the normal line

Using the point-slope form of the equation of a line, the equation of the normal at the point \( (1, 2) \) is:

\[ y - 2 = -1(x - 1) \]

Simplifying this, we get:

\[ y - 2 = -x + 1 \implies y = -x + 3 \]

### Step 4: Find the intersection of the normal with the x-axis

To find where the normal intersects the x-axis, we set \( y = 0 \):

\[ 0 = -x + 3 \implies x = 3 \]

Thus, the normal intersects the x-axis at the point \( (3, 0) \).

### Step 5: Find the equation of the tangent to \( y = e^x \) at the point \( (c, e^c) \)

The slope of the tangent to the curve \( y = e^x \) at the point \( (c, e^c) \) is:

\[ \frac{dy}{dx} = e^x \]

At \( x = c \):

\[ \text{slope} = e^c \]

Using the point-slope form again, the equation of the tangent line is:

\[ y - e^c = e^c(x - c) \]

### Step 6: Find the intersection of the tangent with the x-axis

Setting \( y = 0 \):

\[ 0 - e^c = e^c(x - c) \]

This simplifies to:

\[ -e^c = e^c(x - c) \]

Dividing both sides by \( e^c \) (assuming \( e^c \neq 0 \)):

\[ -1 = x - c \implies x = c - 1 \]

### Step 7: Set the x-coordinates equal

Since both lines intersect at the x-axis, we have:

\[ 3 = c - 1 \]

Solving for \( c \):

\[ c = 4 \]

### Final Answer

Thus, the value of \( c \) is:

\[ \boxed{4} \]