Statement I the equation of the common tangent to the parabolas `y^2=4x` and `x^2=4y` is `x+y+1=0`.
Statement II Both the parabolas are reflected to each other about the line `y=x`.

Options

A. Statement I is true, Statement II is true , Statement II is a correct explanation for statement I.
B. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I.
C. Statement I is true, Statement II is false.
D. Statement I is false,Statement II is true.

Answer

To solve the problem, we need to verify both statements regarding the parabolas \( y^2 = 4x \) and \( x^2 = 4y \).

### Step 1: Finding the common tangent to the parabolas

1. **Equation of the first parabola**: The equation \( y^2 = 4x \) represents a parabola that opens to the right.
2. **Equation of the second parabola**: The equation \( x^2 = 4y \) represents a parabola that opens upwards.

### Step 2: General form of the tangent to the first parabola

The equation of the tangent to the parabola \( y^2 = 4x \) can be expressed as:
\[ y = mx + \frac{1}{m} \]
where \( m \) is the slope of the tangent.

### Step 3: Substitute the tangent equation into the second parabola

To find the common tangent, we substitute \( y = mx + \frac{1}{m} \) into the equation of the second parabola \( x^2 = 4y \):
\[ x^2 = 4\left(mx + \frac{1}{m}\right) \]
This simplifies to:
\[ x^2 = 4mx + \frac{4}{m} \]
Rearranging gives us:
\[ x^2 - 4mx - \frac{4}{m} = 0 \]

### Step 4: Condition for tangency

For the line to be a common tangent, this quadratic equation must have exactly one solution, which occurs when the discriminant is zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = -4m \), and \( c = -\frac{4}{m} \):
\[ D = (-4m)^2 - 4(1)\left(-\frac{4}{m}\right) = 16m^2 + \frac{16}{m} \]
Setting the discriminant to zero:
\[ 16m^2 + \frac{16}{m} = 0 \]
Multiplying through by \( m \) (assuming \( m \neq 0 \)):
\[ 16m^3 + 16 = 0 \]
This simplifies to:
\[ m^3 = -1 \implies m = -1 \]

### Step 5: Finding the equation of the common tangent

Substituting \( m = -1 \) back into the tangent equation:
\[ y = -x - 1 \]
This can be rearranged to:
\[ x + y + 1 = 0 \]
Thus, **Statement I** is true.

### Step 6: Verifying Statement II

To show that the parabolas are reflections of each other about the line \( y = x \):

1. **Reflection of \( y^2 = 4x \)**: If we swap \( x \) and \( y \), we get \( x^2 = 4y \), which is exactly the equation of the second parabola.
2. **Conclusion**: Since reflecting \( y^2 = 4x \) gives \( x^2 = 4y \), we can conclude that both parabolas are indeed reflections of each other about the line \( y = x \).

Thus, **Statement II** is also true.

### Final Conclusion

Both statements are true, but Statement II does not serve as a correct explanation for Statement I. Therefore, the answer is:

**B**: Statement I is true, Statement II is true, but Statement II is not the correct explanation for Statement I.