Consider two curves E, P as `E : x^2/9+y^2/8=1 and P : y^2=4x` on the basis of above answer the following Let m be the slope of a line which is tangent to both E & P then

Answer

To find the slope \( m \) of the line that is tangent to both curves \( E \) and \( P \), we will follow these steps:

### Step 1: Identify the equations of the curves
The equations of the curves are given as:
- Ellipse \( E: \frac{x^2}{9} + \frac{y^2}{8} = 1 \)
- Parabola \( P: y^2 = 4x \)

### Step 2: Write the equation of the tangent line to the ellipse
The equation of the tangent line to the ellipse can be expressed as:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
where \( a^2 = 9 \) and \( b^2 = 8 \). Thus, we have:
\[ y = mx \pm \sqrt{9m^2 + 8} \]

### Step 3: Write the equation of the tangent line to the parabola
The equation of the tangent line to the parabola can be expressed as:
\[ y = mx + \frac{a}{m} \]
where \( a = 1 \) (since \( y^2 = 4x \) can be rewritten as \( x = \frac{y^2}{4} \)). Thus, we have:
\[ y = mx + \frac{1}{m} \]

### Step 4: Set the two tangent line equations equal to each other
Since both equations represent the same line, we can set them equal to each other:
\[ mx \pm \sqrt{9m^2 + 8} = mx + \frac{1}{m} \]

### Step 5: Solve for \( m \)
Subtract \( mx \) from both sides:
\[ \pm \sqrt{9m^2 + 8} = \frac{1}{m} \]
Now, squaring both sides gives us:
\[ 9m^2 + 8 = \frac{1}{m^2} \]
Multiplying through by \( m^2 \) to eliminate the fraction:
\[ 9m^4 + 8m^2 - 1 = 0 \]

### Step 6: Substitute \( t = m^2 \)
Let \( t = m^2 \). The equation becomes:
\[ 9t^2 + 8t - 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 9 \cdot (-1)}}{2 \cdot 9} \]
\[ t = \frac{-8 \pm \sqrt{64 + 36}}{18} \]
\[ t = \frac{-8 \pm \sqrt{100}}{18} \]
\[ t = \frac{-8 \pm 10}{18} \]
Calculating the two possible values:
1. \( t = \frac{2}{18} = \frac{1}{9} \)
2. \( t = \frac{-18}{18} = -1 \) (not valid since \( t = m^2 \) cannot be negative)

### Step 8: Find \( m \)
Since \( t = m^2 \), we have:
\[ m^2 = \frac{1}{9} \implies m = \pm \frac{1}{3} \]

### Conclusion
The slopes of the tangent lines to both curves are:
\[ m = \frac{1}{3} \quad \text{and} \quad m = -\frac{1}{3} \]