Let y^2 = 4ax be a parabola and x^2 + y^2 + 2 | vedclass

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(D) The parabola is $y^2 = 4ax$ and the circle is $x^2 + y^2 + 2bx = 0$.Substituting $y^2 = 4ax$ into the circle equation: $x^2 + 4ax + 2bx = 0$.$x(x

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Both $(A)$ and $(B)$

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(D) The parabola is $y^2 = 4ax$ and the circle is $x^2 + y^2 + 2bx = 0$.Substituting $y^2 = 4ax$ into the circle equation: $x^2 + 4ax + 2bx = 0$.$x(x + 4a + 2b) = 0$.For the curves to touch,the roots must be equal,so $x = 0$ is a double root.This implies $4a + 2b = 0$,or $b = -2a$.For the curves to touch externally,the parabola must open in the direction of the circle's center. The circle $x^2 + 2bx + y^2 = 0$ has center $(-b, 0)$ and radius $|b|$.If $a > 0$,the parabola opens to the right,so the center of the circle must be at a positive $x$-coordinate,meaning $-b > 0$,so $b Actually,for external contact,the curves must lie on opposite sides of the common tangent at the point of contact. Given the geometry,$a$ and $b$ must have opposite signs for the parabola to open towards the circle's center. Thus,$a$ and $b$ must have opposite signs,which is not explicitly listed as a single option. Re-evaluating: if $a > 0, b  0$,they touch. Thus,both $(A)$ and $(B)$ are incorrect as stated,but based on the provided options,the condition for external contact is $b = -2a$.

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 $(1, 3)$	$IV$. $x=9$

Let $y^2 = 4ax$ be a parabola and $x^2 + y^2 + 2bx = 0$ be a circle. If the parabola and the circle touch each other externally,then:

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