`y^2=4x-20` ,Tangent to this parabola at (6,2) is also tangent to `x^2/2+y^2/b=1` then find b

Answer

To solve the problem step by step, we will find the value of \( b \) for the given conditions.

### Step 1: Identify the equation of the parabola
The given parabola is:
\[ y^2 = 4x - 20 \]
This can be rewritten as:
\[ y^2 = 4(x - 5) \]
This indicates that the parabola opens to the right with its vertex at \( (5, 0) \).

### Step 2: Find the equation of the tangent at the point (6, 2)
To find the equation of the tangent line to the parabola at the point \( (6, 2) \), we use the formula for the tangent line to the parabola \( y^2 = 4px \):
\[ yy_1 = 2p(x + x_1) \]
Here, \( p = 1 \) (since \( 4p = 4 \)), \( (x_1, y_1) = (6, 2) \).

Substituting the values:
\[ y(2) = 2(1)(x + 6) \]
This simplifies to:
\[ 2y = 2x + 12 \]
Dividing through by 2 gives:
\[ y = x + 6 \]

### Step 3: Determine the slope of the tangent
From the equation \( y = x + 6 \), we can see that the slope \( m \) of the tangent line is:
\[ m = 1 \]

### Step 4: Write the equation of the tangent to the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{2} + \frac{y^2}{b} = 1 \]
The equation of the tangent to this ellipse at any point can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 + b} \]
where \( a^2 = 2 \) and \( m = 1 \).

Substituting these values into the tangent equation gives:
\[ y = x \pm \sqrt{2(1^2) + b} = x \pm \sqrt{2 + b} \]

### Step 5: Set the tangent equations equal
We know the tangent line from the parabola is \( y = x + 6 \). Therefore, we set the two equations equal:
\[ x + 6 = x + \sqrt{2 + b} \]
This implies:
\[ 6 = \sqrt{2 + b} \]

### Step 6: Solve for \( b \)
To eliminate the square root, we square both sides:
\[ 36 = 2 + b \]
Now, solving for \( b \):
\[ b = 36 - 2 = 34 \]

### Conclusion
Thus, the value of \( b \) is:
\[ \boxed{34} \]