Find the equation of tangent of the curve `y^(2) = 4x+5` which is parallel to the line ` 2x-y=5`.

Answer

To find the equation of the tangent to the curve \( y^2 = 4x + 5 \) that is parallel to the line \( 2x - y = 5 \), we will follow these steps:

### Step 1: Determine the slope of the given line
The equation of the line can be rearranged into slope-intercept form \( y = mx + b \):
\[ 2x - y = 5 \implies y = 2x - 5 \]
From this, we see that the slope \( m \) of the line is \( 2 \).

### Step 2: Find the derivative of the curve
To find the slope of the tangent to the curve, we need to differentiate the curve \( y^2 = 4x + 5 \) implicitly with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x + 5) \]
Using the chain rule on the left side, we get:
\[ 2y \frac{dy}{dx} = 4 \]
Now, solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 3: Set the derivative equal to the slope of the line
Since we want the tangent to be parallel to the line, we set the derivative equal to the slope of the line:
\[ \frac{2}{y} = 2 \]
Solving for \( y \):
\[ 2 = 2y \implies y = 1 \]

### Step 4: Find the corresponding \( x \) value
Now we substitute \( y = 1 \) back into the original curve equation to find \( x \):
\[ y^2 = 4x + 5 \implies 1^2 = 4x + 5 \implies 1 = 4x + 5 \]
Rearranging gives:
\[ 4x = 1 - 5 \implies 4x = -4 \implies x = -1 \]
Thus, the point of tangency is \( (-1, 1) \).

### Step 5: Write the equation of the tangent line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = 2 \), \( x_1 = -1 \), and \( y_1 = 1 \):
\[ y - 1 = 2(x + 1) \]
Expanding this:
\[ y - 1 = 2x + 2 \implies y = 2x + 3 \]

### Final Answer
The equation of the tangent to the curve \( y^2 = 4x + 5 \) that is parallel to the line \( 2x - y = 5 \) is:
\[ \boxed{y = 2x + 3} \]