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

Question

If the equation of the parabola with vertex $V(\frac{3}{2},3)$ 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 :

Accepted Answer

Equation of axis

$\begin{gathered} y-3=2(x-\frac{3}{2}) \ y-2x=0 \end{gathered}$

foot of directrix

$\begin{gathered} \, y-2x=0 \ \, \Rightarrow (0,0) \ 2y+x=0 \end{gathered}$

$\begin{gathered} =(3,6) \ {PS}^{2}={P}^{2} \ (x-3{)}^{2}+(y-6{)}^{2}={(\frac{x+2y}{\sqrt{5}})}^{2} \ 4{x}^{2}+{y}^{2}-4xy-30x-60y+225=0 \ \Rightarrow \alpha =4,\beta =1,\gamma =4\Rightarrow \alpha +\beta +\gamma =9 \end{gathered}$

Solutions · 1