a) Concept used:

The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x-axis is given below:

$\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1:a>b$

The length of the major axis is 2 a and the length of the minor axis is 2 b.

Calculation

Consider the center of the ellipse as (3, 2), the length of the horizontal major axis is 8 and the length of minor axis is 6.

Since the length of the major axis is 8, then $2a=8,\Rightarrow a=4.$

Since the length of the minor axis is 6, then $2b=6,\Rightarrow b=3.$

Hence, the standard equation of the ellipse with center (3, 2) and $a=4,b=3$ is given below:

$\frac{{(x-3)}^{2}}{{4}^{2}}+\frac{{(y-2)}^{2}}{{3}^{2}}=1$

$\frac{{(x-3)}^{2}}{16}+\frac{{(y-2)}^{2}}{9}=1$

(b) Consider the vertices of the hyperbola as (3, 3) and (-3, 3). The foci of the hyperbola are (4, 3) and (-4, 3).

The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis.

Thus, the equation of the hyperbola will have the form $\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1.$

The center is halfway between the vertices (-3, 3) and (3, 3).

Now, Applying the midpoint formula to get center of the hyperbola as below:

$(h,k)=(\frac{3-3}{2},\frac{3+3}{2})=(0,3)$

The length of the transverse axis 2a, is bounded by the vertices.

Hence, the distance between x-coordinates is obtain as follows:

$2a=\left|3-(-3)\right|$

$2a=\left|6\right|=6$

$a=3$

${a}^{2}=9$

The coordinates of the foci are $(h\pm c,k)$.

So, we have $(h-c,k)=(-4,3){\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}(h+c,k)=(4,3).$

Solve the above coordinates for c as follows:

$(h-c,k)=(-4,3)$

$h-c=-4,k=3$

$0-c=-4$

$(h,k)=(0,3)$

$c=4$

Now obtain the value of $b}^{2$

using the relation ${c}^{2}={a}^{2}+{b}^{2}\text{}\text{with}\text{}a=3,c=4$ as follows:

$c}^{2}={a}^{2}+{b}^{2$

$4}^{2}={3}^{2}+{b}^{2$

${b}^{2}=16-9$

${b}^{2}=7$

Hence, the standard equation of the hyperbola with center (0, 3) and transverse axis parallel to the x-axis is given below:

$\frac{{(x-0)}^{2}}{9}-\frac{{(y-3)}^{2}}{7}=1$

(c) Consider the vertex of the parabola as (-2, 4) and the focus as (-4, 4).

Here, the y- coordinates of the vertex and focus are same. Hence, this is a regular horizontal parabola, where y part is square.

Here, take the difference between x coordinate of the focus and vertex as $p=-4+2=-2.$

Hence, the equation of the parabola with vertex (-2, 4) and focus (-4, 4) is given below:

${(y-k)}^{2}=4p(x-h)$