If the equation of the parabola with vertex $V\left(\frac32,3\right)$ 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 :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The foot of the perpendicular from $V$ to the directrix is the origin ($V-\frac32(1,2)=(0,0)$), so the focus is $S=2V-0=(3,6)$. The parabola is $SP^2=\frac{(x+2y)^2}5$: $5\left[(x-3)^2+(y-6)^2\right]=(x+2y)^2\Rightarrow4x^2+y^2-4xy-30x-60y+225=0$. So $\alpha=4,\beta=1,\gamma=4$ and the sum is $9$. Answer (2).

Q. If the equation of the parabola with vertex $V\left(\frac32,3\right)$ 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 :

Explanation

Reviewed by Dr. Abubacker

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