For the given circle 2x^2 + 2y^2 = 5 and the | vedclass

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(C) The circle is $x^2 + y^2 = \frac{5}{2}$. The radius $r = \sqrt{\frac{5}{2}}$.For the parabola $y^2 = 4ax$,where $4a = 4\sqrt{5}$,so $a = \sqrt{5}$

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Statement-$I$ is true,Statement-$II$ is false.

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(C) The circle is $x^2 + y^2 = \frac{5}{2}$. The radius $r = \sqrt{\frac{5}{2}}$.For the parabola $y^2 = 4ax$,where $4a = 4\sqrt{5}$,so $a = \sqrt{5}$.The tangent to the parabola $y^2 = 4\sqrt{5}x$ is $y = mx + \frac{a}{m} = mx + \frac{\sqrt{5}}{m}$.For this to be a tangent to the circle $x^2 + y^2 = \frac{5}{2}$,the perpendicular distance from the center $(0,0)$ to the line $mx - y + \frac{\sqrt{5}}{m} = 0$ must equal the radius $r = \sqrt{\frac{5}{2}}$.So,$\frac{|\frac{\sqrt{5}}{m}|}{\sqrt{m^2 + 1}} = \sqrt{\frac{5}{2}}$.Squaring both sides: $\frac{5/m^2}{m^2 + 1} = \frac{5}{2}$.$\frac{1}{m^2(m^2 + 1)} = \frac{1}{2} \implies m^2(m^2 + 1) = 2 \implies m^4 + m^2 - 2 = 0$.Factoring: $(m^2 + 2)(m^2 - 1) = 0$. Since $m$ is real,$m^2 = 1$,so $m = \pm 1$.For $m = 1$,the tangent is $y = x + \sqrt{5}$. Statement-$I$ is true.For Statement-$II$,the condition derived is $m^4 + m^2 - 2 = 0$,which is different from $m^4 - 3m^2 + 2 = 0$. Thus,Statement-$II$ is false.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

$List-I$	$List-II$
$(I) \ 2h + k$	$(P) \ 6$
$(II) \ \frac{\text{Length of } ZW}{\text{Length of } XY}$	$(Q) \ \sqrt{6}$
$(III) \ \frac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW}$	$(R) \ \frac{5}{4}$
$(IV) \ \alpha$	$(S) \ \frac{21}{5}$
	$(T) \ 2\sqrt{6}$
	$(U) \ \frac{10}{3}$

For the given circle $2x^2 + 2y^2 = 5$ and the parabola $y^2 = 4\sqrt{5}x$:
Statement-$I$: The equation of the common tangent to these curves is $y = x + \sqrt{5}$.
Statement-$II$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is a common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.

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For the given circle $2x^2 + 2y^2 = 5$ and the parabola $y^2 = 4\sqrt{5}x$:Statement-$I$: The equation of the common tangent to these curves is $y = x + \sqrt{5}$.Statement-$II$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is a common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.