If the equation of the parabola with vertex V(32,3) and directrix x+2y=0 is x^2+ y^2- xy-3…

Question

If the equation of the parabola with vertex $V\left(\dfrac32,3\right)$ and directrix $x+2y=0$ is $\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0$, then $\alpha+\beta+\gamma$ is equal to:

Accepted Answer

Axis through $V$ perpendicular to the directrix: slope $2$, giving $y-2x=0$, which meets the directrix at $(0,0)$ (the foot of the directrix). Reflecting $(0,0)$ in $V$ gives the focus $(3,6)$.

Parabola: $\text{(distance to focus)}^2=\text{(distance to directrix)}^2$:

$$(x-3)^2+(y-6)^2=\frac{(x+2y)^2}{5}.$$

$5(x^2-6x+9+y^2-12y+36)=x^2+4xy+4y^2$

$$\Rightarrow 4x^2+y^2-4xy-30x-60y+225=0.$$

So $\alpha=4,\ \beta=1,\ \gamma=4$, and $\alpha+\beta+\gamma=9$.

Solutions · 1