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(A) Any line passing through the origin is given by $y = mx$.Since this line is a tangent to the parabola $y^2 = 4a(x - a)$,substituting $y = mx$ into

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$90$

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(A) Any line passing through the origin is given by $y = mx$.Since this line is a tangent to the parabola $y^2 = 4a(x - a)$,substituting $y = mx$ into the equation gives:$(mx)^2 = 4a(x - a)$$m^2x^2 - 4ax + 4a^2 = 0$For the line to be a tangent,the roots of this quadratic equation must be equal,so the discriminant $D = 0$:$(-4a)^2 - 4(m^2)(4a^2) = 0$$16a^2 - 16a^2m^2 = 0$$16a^2(1 - m^2) = 0$Since $a 
eq 0$,we have $1 - m^2 = 0$,which implies $m^2 = 1$,so $m = 1$ or $m = -1$.The slopes of the two tangents are $m_1 = 1$ and $m_2 = -1$.Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the tangents are perpendicular to each other.Therefore,the angle between the tangents is $90^\circ$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

The angle between the tangents drawn from the origin to the parabola $y^2 = 4a(x - a)$ is ............... $^\circ$.

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