We are given a circle: $2x^2+2y^2=5$ and a parabola: $y^2=4\sqrt{5}x$. We need to analyze the truth of two statements about their common tangents.

Step 1: Simplify the Equations

The circle equation can be simplified by dividing by 2: $x^2+y^2=\frac{5}{2}$. So, center is (0,0) and radius $r=\sqrt{\frac{5}{2}}=\frac{\sqrt{5}}{\sqrt{2}}$.

The parabola is $y^2=4\sqrt{5}x$. Compare with standard form $y^2=4ax$, we get $a=\sqrt{5}$.

Step 2: Condition for Tangent to Parabola

The equation of a tangent to the parabola $y^2=4ax$ with slope m is $y=mx+\frac{a}{m}$. Here, $a=\sqrt{5}$, so the tangent is $y=mx+\frac{\sqrt{5}}{m}$ for $m\ne 0$. This is given in Statement-II.

Step 3: Condition for Tangent to Circle

For the line $y=mx+c$ to be tangent to the circle $x^2+y^2=\frac{5}{2}$, the perpendicular distance from center (0,0) to the line must equal the radius.

Distance $d=\frac{\left|c\right|}{\sqrt{m^2+1}}$. Set $d=r=\sqrt{\frac{5}{2}}$.

So, $\frac{\left|c\right|}{\sqrt{m^2+1}}=\sqrt{\frac{5}{2}}$, which gives $c^2=\frac{5}{2}(m^2+1)$.

Step 4: Apply Both Conditions for Common Tangent

For the line to be common tangent, it must be tangent to both curves. From parabola, we have $c=\frac{\sqrt{5}}{m}$. Substitute into circle condition:

${\left(\frac{\sqrt{5}}{m}\right)}^2=\frac{5}{2}(m^2+1)$

Simplify: $\frac{5}{m^2}=\frac{5}{2}(m^2+1)$

Multiply both sides by $m^2$: $5=\frac{5}{2}{m}^2(m^2+1)$

Divide both sides by 5: $1=\frac{1}{2}{m}^2(m^2+1)$

Multiply both sides by 2: $2=m^2(m^2+1)$

So, $m^4+m^2-2=0$

This is a quadratic in $m^2$: Let $t=m^2$, then $t^2+t-2=0$

Solve: $t=\frac{-1\pm \sqrt{1+8}}{2}=\frac{-1\pm 3}{2}$, so $t=1$ or $t=-2$ (discard negative). Thus, $m^2=1$, so $m=1$ or $m=-1$.

Therefore, the condition is $m^2=1$, not m⁴ - 3m² + 2 = 0 as stated in Statement-II. So, Statement-II is false.

Step 5: Check Statement-I

Statement-I claims that $y=x+\sqrt{5}$ is a common tangent. This line has slope m=1 and intercept c=$\sqrt{5}$.

For parabola: The tangent with m=1 is $y=x+\frac{\sqrt{5}}{1}=x+\sqrt{5}$, which matches.

For circle: Check distance from center to line: $d=\frac{\left|\sqrt{5}\right|}{\sqrt{1+1}}=\frac{\sqrt{5}}{\sqrt{2}}=r$. So, it is tangent to the circle.

Thus, Statement-I is true.

Conclusion

Statement-I is true, but Statement-II is false because it gives an incorrect condition (m⁴ - 3m² + 2 = 0) instead of the correct condition (m² = 1).

Related Topics and Formulae

Tangent to Parabola: For $y^2=4ax$, tangent with slope m is $y=mx+\frac{a}{m}$.

Tangent to Circle: For circle $x^2+y^2=r^2$, tangent with slope m is $y=mx\pm r\sqrt{m^2+1}$.

Common Tangent: A line that is tangent to two or more curves. Conditions from each curve must be satisfied simultaneously.