Given : A circle, `2x^2+""2y^2=""5` and a parabola, `y^2=""4sqrt(5)""x` . Statement - I : An equation of a common tangent to these curves is `y="x+"sqrt(5)` Statement - II : If the line, `y=m x+(sqrt(5))/m(m!=0)` is their common tangent, then m satisfies `m^4-3m^2+""2""=0.` (1) Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I (2) Statement -I is True; Statement -II is False. (3) Statement -I is False; Statement -II is True (4) Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I

Options

A. Statement1 is correct, statement 2 is correct, statement 2 is a correct explanation for statement 1
B. Statement1 is correct, statement 2 is correct, Statement 2 is not a correct explanation for statement 1
C. Statement1 is correct, statement 2 is incorrect
D. Statement1 is incorrect, statement 2 is correct

Answer

To solve the problem, we need to analyze the given statements about the common tangent to a circle and a parabola.

### Step-by-Step Solution:

1. **Identify the Equations**:
- The equation of the circle is given as:
\[ 2x^2 + 2y^2 = 5 \]
This can be simplified to:
\[ x^2 + y^2 = \frac{5}{2} \]
which represents a circle centered at the origin with radius \(\sqrt{\frac{5}{2}}\).

- The equation of the parabola is given as:
\[ y^2 = 4\sqrt{5}x \]
This is a standard form of a parabola that opens to the right.

2. **Common Tangent Form**:
- The general form of the equation of a tangent to the parabola \(y^2 = 4ax\) is:
\[ y = mx + \frac{a}{m} \]
where \(a = \sqrt{5}\) in our case. Therefore, the equation of the tangent becomes:
\[ y = mx + \frac{\sqrt{5}}{m} \]

3. **Substituting into the Circle's Equation**:
- We substitute \(y = mx + \frac{\sqrt{5}}{m}\) into the circle's equation:
\[ x^2 + \left(mx + \frac{\sqrt{5}}{m}\right)^2 = \frac{5}{2} \]
- Expanding this gives:
\[ x^2 + \left(m^2x^2 + 2mx\frac{\sqrt{5}}{m} + \frac{5}{m^2}\right) = \frac{5}{2} \]
- Simplifying:
\[ (1 + m^2)x^2 + 2\sqrt{5}x + \left(\frac{5}{m^2} - \frac{5}{2}\right) = 0 \]

4. **Finding the Discriminant**:
- For the line to be a tangent, the discriminant of this quadratic equation must be zero:
\[ b^2 - 4ac = 0 \]
- Here, \(b = 2\sqrt{5}\), \(a = 1 + m^2\), and \(c = \frac{5}{m^2} - \frac{5}{2}\).
- Setting up the discriminant:
\[ (2\sqrt{5})^2 - 4(1 + m^2)\left(\frac{5}{m^2} - \frac{5}{2}\right) = 0 \]

5. **Solving the Discriminant Equation**:
- Calculate:
\[ 20 - 4(1 + m^2)\left(\frac{5}{m^2} - \frac{5}{2}\right) = 0 \]
- This leads to:
\[ 20 - 20(1 + m^2) + 10m^2 = 0 \]
- Rearranging gives:
\[ m^4 - 3m^2 + 2 = 0 \]

6. **Analyzing the Statements**:
- **Statement I**: The equation \(y = x + \sqrt{5}\) is indeed a common tangent, as we derived the general form and found that it fits.
- **Statement II**: The derived equation \(m^4 - 3m^2 + 2 = 0\) is correct and shows the conditions for \(m\).

### Conclusion:
- **Statement I is True**.
- **Statement II is True** and is a correct explanation for Statement I.

Thus, the correct option is:
**(4) Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I.**