Animated Solution for Mathematics - Conic Sections: 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 :

Welcome, future engineers! Today, we are going to peel back the layers of a classic coordinate geometry problem. We are given the vertex of a parabola and its directrix, and our mission is to reconstruct the entire equation.

It might look like a daunting algebraic mess at first, but I promise you, once you see the geometry, the math will flow like water.

Imagine standing on the coordinate plane. You have a directrix line $x+2y=0$ and a vertex $V(\frac{3}{2},3)$.

The first thing we need is the axis of symmetry. We know the axis is perpendicular to the directrix. Since the directrix has a slope of $-\frac{1}{2}$, our axis must have a slope of $2$.

Using the point-slope form with our vertex, we get the equation $y-3=2(x-\frac{3}{2})$, which simplifies beautifully to $2x-y=0$. This line is our path forward.

Now, let's find the intersection point $Z$ where the axis meets the directrix. Solving the system $x+2y=0$ and $2x-y=0$ is straightforward substitution.

If $y=2x$, then $x+2(2x)=0$, which gives us $x=0$ and $y=0$. So, the intersection point $Z$ is the origin $(0,0)$.

Here comes the 'JEE magic' moment: the vertex $V$ is the midpoint of the segment connecting the intersection point $Z$ and the focus $S(h,k)$. Using the midpoint formula:

We instantly find our focus $S$ at $(3,6)$.

Now that we have the focus $S(3,6)$ and the directrix $x+2y=0$, we invoke the definition of a parabola: for any point $P(x,y)$ on the curve, the distance to the focus $PS$ must equal the perpendicular distance to the directrix $PM$.

To avoid the headache of square roots, we square both sides: $P{S}^2=P{M}^2$. The distance formula gives us $(x-3{)}^2+(y-6{)}^2$ on the left. The perpendicular distance formula gives us the following on the right:

This is where precision matters. We have the following equality:

Expanding the left side gives $x^2-6x+9+y^2-12y+36$. Multiplying by $5$ and setting it equal to the expansion of the right side, $x^2+4y^2+4xy$, we get:

Rearranging everything to one side, we arrive at:

Comparing this to the given form $\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0$, we identify $\alpha =4$, $\beta =1$, and $\gamma =4$.

The sum $\alpha +\beta +\gamma$ is $4+1+4=9$.

You have just conquered a complex conic section problem by breaking it down into simple, logical steps. Keep this mindset, and no problem will ever be too difficult for you!