3. Find a Cartesian equation for the hyperbola centered at the origin that has focus at $$ (3,0) $$ and the line $$ x=1 $$ as the corresponding directrix.

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1. We start with the definition of a conic section in terms of a focus and directrix. For any point $$ (x,y) $$ on the conic, the distance to the focus is $$ e $$ times the distance to the directrix. Here the focus is $$ (3,0) $$ and the corresponding directrix is the line $$ x=1 $$. Thus,
   
   $$
   \sqrt{(x-3)^2+y^2} = e\, |x-1|.
   $$

2. Since the hyperbola is centered at the origin and one focus is $$ (3,0) $$, by symmetry the other focus must be at $$ (-3,0) $$. Therefore, the hyperbola has the standard form

$$
   \frac{x^2}{a^2} - \frac{y^2}{b^2} =1,
   $$
   
   with vertices at $$ (\pm a, 0) $$ and foci at $$ (\pm c,0) $$ where $$ c =3 $$.

3. For a hyperbola, the eccentricity is given by

$$
   e = \frac{c}{a}.
   $$
   
   Hence, we have

$$
   e=\frac{3}{a}.
   $$
   
4. The definition of the conic also implies that any point on the hyperbola satisfies

$$
   \frac{\text{distance to focus}}{\text{distance to directrix}} = e.
   $$
   
   Consider the vertex on the right branch, $$ (a,0) $$. Its distance to the focus $$ (3,0) $$ is

$$
   |3-a|,
   $$
   
   and its distance to the directrix $$ x=1 $$ is

$$
   |a-1|.
   $$
   
   Therefore, at the vertex,
   
   $$
   \frac{3-a}{a-1} = e.
   $$

5. Substitute the expression $$ e = \frac{3}{a} $$ obtained earlier into the above relation:

$$
   \frac{3-a}{a-1} = \frac{3}{a}.
   $$
   
   Solve for $$ a $$ by cross-multiplying:
   
   $$
   3(a-1)=3a-(a)(a) = a(3-a).
   $$
   
   Expanding the left-hand side:
   
   $$
   3a-3 = 3a - a^2.
   $$
   
   Subtract $$ 3a $$ from both sides:
   
   $$
   -3 = -a^2,
   $$
   
   which gives
   
   $$
   a^2 = 3 \quad \Longrightarrow \quad a = \sqrt{3}.
   $$

6. Now, substitute $$ a = \sqrt{3} $$ back into the expression for eccentricity:

$$
   e = \frac{3}{\sqrt{3}}=\sqrt{3}.
   $$

7. For a hyperbola, the relationship between $$ a $$, $$ b $$, and $$ c $$ is

$$
   c^2 = a^2+b^2.
   $$
   
   Substitute $$ c=3 $$ and $$ a^2=3 $$:
   
   $$
   9 = 3+b^2,
   $$
   
   so
   
   $$
   b^2=6.
   $$

8. The Cartesian equation of the hyperbola is then

$$
   \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,
   $$
   
   with $$ a^2=3 $$ and $$ b^2=6 $$. Thus, the equation becomes

$$
   \frac{x^2}{3} - \frac{y^2}{6} = 1.
   $$

9. As a check, the directrix corresponding to the right branch of a hyperbola with eccentricity $$ e $$ is given by

$$
   x = \frac{a}{e} = \frac{\sqrt{3}}{\sqrt{3}} = 1,
   $$
   
   which matches the given directrix.

The Cartesian equation for the hyperbola is

$$
\frac{x^2}{3} - \frac{y^2}{6} = 1.
$$
The Cartesian equation of the hyperbola is $$ \frac{x^2}{3} - \frac{y^2}{6} = 1 $$.

Find a Cartesian equation for the hyperbola centered at the
origin that has focus at and the line as the
corresponding directrix.

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Find a Cartesian equation for the hyperbola centered at the origin that has focus at and the line as the corresponding directrix.