Write the equation of each conic section, given the following characteristics: a) Write the equation of an ellipse with center at (3, 2) and horizonta

The following equation for an ellipse with a major axis parallel to the x-axis and a center at (h, k) is presented in standard form.
${\displaystyle \frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1:a>b}$ 
The major axis is 2 a long, while the minor axis is 2 b long.
Calculation 
Consider that the ellipse's center is at (3, 2), that the major horizontal axis is 8 lengths long, and that the minor axis is 6 lengths long.
Since the length of the major axis is 8, then ${\displaystyle 2a=8,\Rightarrow a=4.}$ 
Since the length of the minor axis is 6, then ${\displaystyle 2b=6,\Rightarrow b=3.}$ 
Consequently, the ellipse's basic equation (3, 2) and ${\displaystyle a=4,b=3}$ is given below: 
${\displaystyle \frac{{(x-3)}^2}{4^2}+\frac{{(y-2)}^2}{3^2}=1}$ 
${\displaystyle \frac{{(x-3)}^2}{16}+\frac{{(y-2)}^2}{9}=1}$ 
(b) Assume that the hyperbola's vertices are (3, 3) and (-3, 3). The hyperbola's foci are (4, 3), and (-4, 3).
The transverse axis is parallel to the x-axis because the y-coordinates of the vertices and foci are the same.
Consequently, the hyperbola's equation will take the form ${\displaystyle \frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1.}$ 
Halfway between the vertices (-3, 3) and is the center (3, 3).
Applying the midpoint formula now will yield the hyperbola's center, as seen below:
${\displaystyle (h,k)=(\frac{3-3}{2},\frac{3+3}{2})=(0,3)}$ 
The vertices define the length of the transverse axis 2a.
Therefore, the following formula is used to find the distance between x-coordinates:
${\displaystyle 2a=\left|3-(-3)\right|}$ 
${\displaystyle 2a=\left|6\right|=6}$ 
${\displaystyle a=3}$ 
${\displaystyle a^2=9}$ 
The coordinates of the foci are ${\displaystyle (h\pm c,k)}$. 
So, we have ${\displaystyle (h-c,k)=(-4,3)\text{and}(h+c,k)=(4,3).}$ 
The aforementioned coordinates for c should be solved as follows:
${\displaystyle (h-c,k)=(-4,3)}$ 
${\displaystyle h-c=-4,k=3}$ 
${\displaystyle 0-c=-4}$ 
${\displaystyle (h,k)=(0,3)}$ 
${\displaystyle c=4}$ 
Now obtain the value of ${\displaystyle b^2}$ 
using the relation ${\displaystyle c^2=a^2+b^2\text{with}a=3,c=4}$ as follows: 
${\displaystyle c^2=a^2+b^2}$ 
${\displaystyle 4^2=3^2+b^2}$ 
${\displaystyle b^2=16-9}$ 
${\displaystyle b^2=7}$ 
As a result, the following is the equation for the hyperbola's standard form with its center at (0, 3) and its transverse axis parallel to the x-axis:
${\displaystyle \frac{{(x-0)}^2}{9}-\frac{{(y-3)}^2}{7}=1}$ 
(c) Assume that the focus is at and the vertex of the parabola is at (-2, 4) (-4, 4).
The focus and vertex here have the same y-coordinate. So, this is a typical horizontal parabola with a square y portion.
Take the difference between the focus's x coordinate and the vertex's as ${\displaystyle p=-4+2=-2.}$ 
As a result, the following equation for the parabola with vertex (-2, 4) and focus (-4, 4) is provided:
${\displaystyle {(y-k)}^2=4p(x-h)}$