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

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Aicharya · Accepted Answer

$\begin{aligned} & 	herefore \quad 	ext { Equation of parabola is } P S=(1) P M \ & \Rightarrow P S^2=P M^2 \ & (x-3)^2+(y-6)^2=\left(\frac{x+2 y}{\sqrt{5}}ight)^2 \ & 5 x^2-30 x+5 y^2-60 y+225=x^2+4 y^2+4 x y \ & 4 x^2+y^2-4 x y-30 x-60 y+225=0 \end{aligned}$ We get : $\alpha=4, \beta=1, \gamma=4$ $	herefore \alpha+\beta+\gamma=9$ ,

If the equation of the parabola with vertex $\mathrm{V}\left(\frac{3}{2}, 3\right)$ and the directrix $x+2…

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