Let y=x+1 is axis of parabola, y+x-4=0 is tangent of same parabola at its vertex and y=2x+3 is one of its tangents. Then find the focus of the parabola.

Explanation

To find the focus of the parabola given the conditions, we can follow these steps:

### Step 1: Identify the Axis of the Parabola
The axis of the parabola is given as \( y = x + 1 \). This means that the parabola opens either to the right or to the left.

### Step 2: Identify the Tangent at the Vertex
The tangent at the vertex is given by the equation \( y + x - 4 = 0 \). We can rewrite this as:
\[ y = -x + 4 \]

### Step 3: Find the Vertex of the Parabola
Since the axis of the parabola is \( y = x + 1 \), the vertex of the parabola lies on this line. Let's denote the vertex as \( (h, k) \). Since the vertex lies on the axis, we can express \( k \) in terms of \( h \):
\[ k = h + 1 \]

### Step 4: Set Up the Equation of the Tangent
The tangent line at the vertex must also satisfy the condition of being perpendicular to the axis of the parabola. The slope of the line \( y = -x + 4 \) is -1. The slope of the axis \( y = x + 1 \) is 1. Since the slopes are negative reciprocals, this confirms that the line is indeed a tangent at the vertex.

### Step 5: Find the Intersection of the Tangents
We also have another tangent line given by \( y = 2x + 3 \). To find the point of intersection of the two tangent lines \( y = -x + 4 \) and \( y = 2x + 3 \), we set them equal to each other:
\[ -x + 4 = 2x + 3 \]
Solving for \( x \):
\[ 4 - 3 = 2x + x \implies 1 = 3x \implies x = \frac{1}{3} \]
Now, substituting \( x = \frac{1}{3} \) back into one of the equations to find \( y \):
\[ y = 2\left(\frac{1}{3}\right) + 3 = \frac{2}{3} + 3 = \frac{2}{3} + \frac{9}{3} = \frac{11}{3} \]
Thus, the point of intersection is \( \left(\frac{1}{3}, \frac{11}{3}\right) \).

### Step 6: Calculate the Slope of the Line from the Vertex to the Intersection Point
Now, we calculate the slope \( m_2 \) from the vertex \( (h, k) \) to the intersection point \( \left(\frac{1}{3}, \frac{11}{3}\right) \):
\[ m_2 = \frac{k - \frac{11}{3}}{h - \frac{1}{3}} \]
Using the condition that the product of the slopes \( m_1 \cdot m_2 = -1 \) (where \( m_1 = 2 \)):
\[ 2 \cdot m_2 = -1 \implies m_2 = -\frac{1}{2} \]

### Step 7: Set Up the Equation
Substituting \( m_2 \) into the slope formula gives:
\[ \frac{k - \frac{11}{3}}{h - \frac{1}{3}} = -\frac{1}{2} \]
Cross-multiplying:
\[ 2(k - \frac{11}{3}) = -1(h - \frac{1}{3}) \]
Expanding and simplifying:
\[ 2k - \frac{22}{3} = -h + \frac{1}{3} \]
Rearranging gives:
\[ 2k + h = \frac{22}{3} + \frac{1}{3} = \frac{23}{3} \]

### Step 8: Substitute for \( k \)
From Step 3, we have \( k = h + 1 \). Substitute this into the equation:
\[ 2(h + 1) + h = \frac{23}{3} \]
This simplifies to:
\[ 3h + 2 = \frac{23}{3} \]
Multiplying through by 3:
\[ 9h + 6 = 23 \implies 9h = 17 \implies h = \frac{17}{9} \]

### Step 9: Find \( k \)
Substituting back to find \( k \):
\[ k = h + 1 = \frac{17}{9} + 1 = \frac{17}{9} + \frac{9}{9} = \frac{26}{9} \]

### Step 10: Focus of the Parabola
The focus of the parabola is given by \( (h, k) \):
\[ \text{Focus} = \left(\frac{17}{9}, \frac{26}{9}\right) \]

### Final Answer
The focus of the parabola is \( \left(\frac{17}{9}, \frac{26}{9}\right) \).