A line passing through the point A(-2, 0) tou | vedclass

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(C) Let the equation of the line passing through $A(-2, 0)$ be $y = m(x + 2)$.Substituting $x = \frac{y}{m} - 2$ into the parabola equation $y^2 = x -

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$\frac{8}{3}$

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(C) Let the equation of the line passing through $A(-2, 0)$ be $y = m(x + 2)$.Substituting $x = \frac{y}{m} - 2$ into the parabola equation $y^2 = x - 2$,we get $y^2 = \frac{y}{m} - 2 - 2$,which simplifies to $y^2 - \frac{y}{m} + 4 = 0$,or $my^2 - y + 4m = 0$.Since the line is tangent to the parabola,the discriminant $D = 0$.$(-1)^2 - 4(m)(4m) = 0 \implies 1 - 16m^2 = 0 \implies m^2 = \frac{1}{16} \implies m = \frac{1}{4}$ (since $B$ is in the first quadrant,$m > 0$).The equation of the tangent is $y = \frac{1}{4}(x + 2)$,or $x = 4y - 2$.The point of tangency $B$ is found by substituting $m = \frac{1}{4}$ into $y^2 - \frac{y}{m} + 4 = 0$,giving $y^2 - 4y + 4 = 0$,so $(y - 2)^2 = 0$,which means $y = 2$. Then $x = 4(2) - 2 = 6$. So $B = (6, 2)$.The area bounded by the line $AB$,the parabola $P$,and the $x$-axis is given by integrating with respect to $y$ from $y = 0$ to $y = 2$:Area $= \int_{0}^{2} (x_{	ext{line}} - x_{	ext{parabola}}) dy = \int_{0}^{2} ((4y - 2) - (y^2 + 2)) dy = \int_{0}^{2} (4y - 4 - y^2) dy$.Area $= [2y^2 - 4y - \frac{y^3}{3}]_{0}^{2} = (2(4) - 4(2) - \frac{8}{3}) - 0 = 8 - 8 - \frac{8}{3} = |-\frac{8}{3}| = \frac{8}{3}$ square units.

$A$ line passing through the point $A(-2, 0)$ touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$,the parabola $P$,and the $x$-axis is:

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