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(A) The equation of the parabola is $y^{2} = 8x$,so $4a = 8 \Rightarrow a = 2$. The directrix is $x = -a = -2$.$1$. Equation of the tangent at $P(2, -

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$-16$

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(A) The equation of the parabola is $y^{2} = 8x$,so $4a = 8 \Rightarrow a = 2$. The directrix is $x = -a = -2$.$1$. Equation of the tangent at $P(2, -4)$:Using $T = 0$,we have $y(-4) = 4(x + 2)$ $\Rightarrow -4y = 4x + 8$ $\Rightarrow x + y + 2 = 0$.Intersection with directrix $x = -2$: $-2 + y + 2 = 0 \Rightarrow y = 0$. So,$A(-2, 0)$.$2$. Equation of the normal at $P(2, -4)$:The slope of the tangent is $m_{T} = -1$. The slope of the normal is $m_{N} = -1 / m_{T} = 1$.Equation: $y - (-4) = 1(x - 2)$ $\Rightarrow y + 4 = x - 2$ $\Rightarrow x - y - 6 = 0$.Intersection with directrix $x = -2$: $-2 - y - 6 = 0 \Rightarrow y = -8$. So,$B(-2, -8)$.$3$. Since $AQBP$ is a square,the diagonals $AB$ and $PQ$ bisect each other at the same midpoint $M$.Midpoint of $AB = ((-2 + -2) / 2, (0 + -8) / 2) = (-2, -4)$.Midpoint of $PQ = ((a + 2) / 2, (b - 4) / 2) = (-2, -4)$.Equating coordinates:$(a + 2) / 2 = -2$ $\Rightarrow a + 2 = -4$ $\Rightarrow a = -6$.$(b - 4) / 2 = -4$ $\Rightarrow b - 4 = -8$ $\Rightarrow b = -4$.$4$. Calculate $2a + b$:$2(-6) + (-4) = -12 - 4 = -16$.

$A$ tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^{2} = 8x$,which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square,then $2a + b$ is equal to:

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