A tangent line L is drawn at the point (2, -4 | vedclass

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(D) The equation of the tangent to the parabola $y^{2} = 4Ax$ at point $(x_{1}, y_{1})$ is $yy_{1} = 2A(x + x_{1})$.Here,$4A = 8$,so $A = 2$.At point

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(D) The equation of the tangent to the parabola $y^{2} = 4Ax$ at point $(x_{1}, y_{1})$ is $yy_{1} = 2A(x + x_{1})$.Here,$4A = 8$,so $A = 2$.At point $(2, -4)$,the tangent equation is $y(-4) = 4(x + 2)$.$-4y = 4x + 8 \Rightarrow x + y + 2 = 0$.This line $x + y + 2 = 0$ is also tangent to the circle $x^{2} + y^{2} = a$.The perpendicular distance from the center $(0, 0)$ to the line $x + y + 2 = 0$ must equal the radius $\sqrt{a}$.Distance $d = \frac{|0 + 0 + 2|}{\sqrt{1^{2} + 1^{2}}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Since $d = \sqrt{a}$,we have $\sqrt{a} = \sqrt{2}$,which implies $a = 2$.

$A$ tangent line $L$ is drawn at the point $(2, -4)$ on the parabola $y^{2} = 8x$. If the line $L$ is also tangent to the circle $x^{2} + y^{2} = a$,then $a$ is equal to .... .

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