If the equation of the parabola with vertex V | vedclass

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(D) The axis of the parabola is perpendicular to the directrix $x + 2y = 0$ and passes through the vertex $V \left(\frac{3}{2}, 3\right)$.Thus,the slo

Vedclass · Accepted Answer

$9$

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(D) The axis of the parabola is perpendicular to the directrix $x + 2y = 0$ and passes through the vertex $V \left(\frac{3}{2}, 3\right)$.Thus,the slope of the axis is $2$. The equation of the axis is $y - 3 = 2 \left(x - \frac{3}{2}\right)$,which simplifies to $y = 2x$.The foot of the directrix is the intersection of $x + 2y = 0$ and $y = 2x$,which is $(0, 0)$.Since the vertex is the midpoint of the focus $S$ and the foot of the directrix $(0, 0)$,we have $\frac{x_S + 0}{2} = \frac{3}{2}$ and $\frac{y_S + 0}{2} = 3$,so the focus $S$ is $(3, 6)$.Using the definition of a parabola,$PS^2 = PM^2$,where $P(x, y)$ is a point on the parabola and $PM$ is the perpendicular distance to the directrix:$(x - 3)^2 + (y - 6)^2 = \left(\frac{x + 2y}{\sqrt{5}}\right)^2$$5(x^2 - 6x + 9 + y^2 - 12y + 36) = x^2 + 4y^2 + 4xy$$5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4y^2 + 4xy$$4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$Comparing with $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$,we get $\alpha = 4, \beta = 1, \gamma = 4$.Therefore,$\alpha + \beta + \gamma = 4 + 1 + 4 = 9$.

If the equation of the parabola with vertex $V \left(\frac{3}{2}, 3\right)$ and the directrix $x + 2y = 0$ is $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$,then $\alpha + \beta + \gamma$ is equal to:

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