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 :
Solution
<p>Equation of axis</p>
<p>$$\begin{aligned}
& y-3=2\left(x-\frac{3}{2}\right) \\
& y-2 x=0
\end{aligned}$$</p>
<p>foot of directrix</p>
<p>$$\begin{aligned}
& \quad y-2 x=0 \\
& \& \quad \Rightarrow(0,0)\\
& 2 y+x=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Focus }=(3,6) \\
& \operatorname{PS}^2=P^2 \\
& (x-3)^2+(y-6)^2=\left(\frac{x+2 y}{\sqrt{5}}\right)^2 \\
& 4 x^2+y^2-4 x y-30 x-60 y+225=0 \\
& \Rightarrow \alpha=4, \beta=1, \gamma=4 \Rightarrow \alpha+\beta+\gamma=9
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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