Animated Solution for Mathematics - Conic Sections: If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3x^2-y^2-\alpha x+\beta y+\gamma =0$, then $\alpha +\beta +\gamma$ is equal to _____.

A Journey of Precision

Welcome, future engineer. Today, we are not just solving for constants $\alpha$, $\beta$, and $\gamma$; we are peeling back the layers of a hyperbola to understand its fundamental structure.

When you see a problem like this, do not panic at the sight of the polynomial $3x^2-y^2-\alpha x+\beta y+\gamma =0$. Instead, see it as a puzzle waiting to be assembled.

The Geometric Foundation

Every hyperbola is defined by its foci. We are given $S^{\prime }(4,2)$ and $S(8,2)$.

The moment you see that the $y$-coordinates are identical, you should feel a sense of relief. This is the universe telling you that the transverse axis is horizontal, lying on the line $y=2$.

The center of the hyperbola, $C(h,k)$, is the midpoint of these two foci. Calculating this is straightforward:

We have our center at $(6,2)$. Now, consider the distance between the foci, which is $2ae$.

Since the distance between $4$ and $8$ is $4$, we have $2ae=4$, which simplifies to $ae=2$, or: This is our first major anchor point.

The Algebraic Bridge

Now, let us look at the equation $3x^2-y^2-\alpha x+\beta y+\gamma =0$. We know the standard form for a horizontal hyperbola is:

If we expand this, the coefficients of $x^2$ and $y^2$ are $\frac{1}{a^2}$ and $-\frac{1}{b^2}$. In our given equation, the coefficients are $3$ and $-1$.

By comparing these ratios, we establish a beautiful relationship:

This is the key that unlocks the door. We now have two equations: $a^2{e}^2=4$ and $b^2=3a^2$.

We also know the fundamental hyperbola identity $b^2=a^2(e^2-1)=a^2{e}^2-a^2$. Substituting our known values, we get:

Solving this gives $4a^2=4$, so $a^2=1$. Consequently, $b^2=3(1)=3$.

The Final Assembly

With $a^2=1$, $b^2=3$, and center $(6,2)$, we can write the standard equation:

To match the form of our given equation, we multiply by $3$:

Expanding these squares is where most students stumble, so let us be precise. We have:

Distributing the $3$, we get $3x^2-36x+108-y^2+4y-4=3$. Bringing everything to one side, we arrive at:

Comparing this to the original equation, we identify $\alpha =36$, $\beta =4$, and $\gamma =101$.

The final sum is:

You have successfully navigated the geometry and the algebra. Take a moment to appreciate the elegance of how these numbers align. You are ready for the next challenge.