SOLUTION: solve for the equation of the hyperbola in general form with directrices x=-5 and x=7, and one focus at (13,5).

Question 1165937: solve for the equation of the hyperbola in general form with directrices x=-5 and x=7, and one focus at (13,5).
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
To find the equation of the hyperbola in general form, we first need to determine the **center**, the distance from the center to the **focus ($c$)**, and the distance from the center to the **directrix ($a/e$)**.
Since the directrices are vertical lines ($x=-5$ and $x=7$), the hyperbola is **horizontal**, meaning its transverse axis is parallel to the $x$-axis.
## 1. Determine Key Parameters
### A. Center ($h, k$)
The center ($h, k$) lies exactly halfway between the two directrices. Since the directrices are defined by $x=-5$ and $x=7$, the $x$-coordinate of the center ($h$) is their average. The focus given is $(13, 5)$, so the $y$-coordinate of the center ($k$) must be 5.
$$h = \frac{-5 + 7}{2} = \frac{2}{2} = 1$$
$$\text{Center} (h, k) = \mathbf{(1, 5)}$$
### B. Distance to Focus ($c$)
The focus is at $(13, 5)$ and the center is at $(1, 5)$. The distance $c$ is the distance between them:
$$c = 13 - 1 = \mathbf{12}$$
### C. Distance between Directrices
The distance between the directrices is $7 - (-5) = 12$. The general distance between the two directrices of a hyperbola is given by $2a/e$.
$$2\frac{a}{e} = 12 \implies \frac{a}{e} = \mathbf{6} \quad \text{(Equation 1)}$$
### D. Eccentricity ($e$) and Semi-transverse Axis ($a$)
We use the relationship between $a, c,$ and the eccentricity $e$:
$$c = a \cdot e$$
Substituting $c=12$:
$$12 = a \cdot e \quad \text{(Equation 2)}$$
Now, we solve the system of equations (1) and (2) for $a$ and $e$:
1. $a = 6e$ (from Eq. 1)
2. $12 = (6e)e$ (substitute into Eq. 2)
$$12 = 6e^2$$
$$e^2 = 2 \implies e = \sqrt{2}$$
Now find $a$:
$$a = 6e = 6\sqrt{2}$$
$$a^2 = (6\sqrt{2})^2 = 36 \times 2 = \mathbf{72}$$
### E. Semi-conjugate Axis ($b^2$)
Use the standard hyperbolic relation $c^2 = a^2 + b^2$.
$$12^2 = 72 + b^2$$
$$144 = 72 + b^2$$
$$b^2 = 144 - 72 = \mathbf{72}$$
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## 2. Standard Form of the Equation
Since the hyperbola is horizontal, the standard form is:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
Substitute $h=1, k=5, a^2=72,$ and $b^2=72$:
$$\frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} = 1$$
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## 3. General Form of the Equation
To convert to the general form ($Ax^2 + Bx + Cy^2 + Dy + E = 0$), multiply the entire equation by the common denominator, 72:
$$72 \left( \frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} \right) = 1 \times 72$$
$$(x-1)^2 - (y-5)^2 = 72$$
Expand the squared terms:
$$(x^2 - 2x + 1) - (y^2 - 10y + 25) = 72$$
Remove parentheses and rearrange:
$$x^2 - 2x + 1 - y^2 + 10y - 25 = 72$$
$$x^2 - y^2 - 2x + 10y - 24 = 72$$
Move the constant term to the left side:
$$x^2 - y^2 - 2x + 10y - 24 - 72 = 0$$
$$x^2 - y^2 - 2x + 10y - 96 = 0$$
The equation of the hyperbola in general form is:
$$\mathbf{x^2 - y^2 - 2x + 10y - 96 = 0}$$
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