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:

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

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 \(\displaystyle{2}{a}={8},\Rightarrow{a}={4}.\)

Since the length of the minor axis is 6, then \(\displaystyle{2}{b}={6},\Rightarrow{b}={3}.\)

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

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

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

(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 \(\displaystyle\frac{{{\left({x}-{h}\right)}^{2}}}{{a}^{2}}-\frac{{{\left({y}-{k}\right)}^{2}}}{{b}^{2}}={1}.\)

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:

\(\displaystyle{\left({h},{k}\right)}={\left(\frac{{{3}-{3}}}{{2}},\frac{{{3}+{3}}}{{2}}\right)}={\left({0},{3}\right)}\)

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

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

\(\displaystyle{2}{a}={\left|{{3}-{\left(-{3}\right)}}\right|}\)

\(\displaystyle{2}{a}={\left|{{6}}\right|}={6}\)

\(\displaystyle{a}={3}\)

\(\displaystyle{a}^{2}={9}\)

The coordinates of the foci are \(\displaystyle{\left({h}\pm{c},{k}\right)}\).

So, we have \(\displaystyle{\left({h}-{c},{k}\right)}={\left(-{4},{3}\right)}{\quad\text{and}\quad}{\left({h}+{c},{k}\right)}={\left({4},{3}\right)}.\)

Solve the above coordinates for c as follows:

\(\displaystyle{\left({h}-{c},{k}\right)}={\left(-{4},{3}\right)}\)

\(\displaystyle{h}-{c}=-{4},{k}={3}\)

\(\displaystyle{0}-{c}=-{4}\)

\(\displaystyle{\left({h},{k}\right)}={\left({0},{3}\right)}\)

\(\displaystyle{c}={4}\)

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:

\(\displaystyle{c}^{2}={a}^{2}+{b}^{2}\)

\(\displaystyle{4}^{2}={3}^{2}+{b}^{2}\)

\(\displaystyle{b}^{2}={16}-{9}\)

\(\displaystyle{b}^{2}={7}\)

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

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

(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 \(\displaystyle{p}=-{4}+{2}=-{2}.\)

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

\(\displaystyle{\left({y}-{k}\right)}^{2}={4}{p}{\left({x}-{h}\right)}\)

\(\displaystyle{\left({y}-{4}\right)}^{2}={4}{\left(-{2}\right)}{\left({x}+{2}\right)}\)

\(\displaystyle{\left({y}-{4}\right)}^{2}=-{8}{\left({x}+{2}\right)}\)

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

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

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 \(\displaystyle{2}{a}={8},\Rightarrow{a}={4}.\)

Since the length of the minor axis is 6, then \(\displaystyle{2}{b}={6},\Rightarrow{b}={3}.\)

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

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

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

(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 \(\displaystyle\frac{{{\left({x}-{h}\right)}^{2}}}{{a}^{2}}-\frac{{{\left({y}-{k}\right)}^{2}}}{{b}^{2}}={1}.\)

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:

\(\displaystyle{\left({h},{k}\right)}={\left(\frac{{{3}-{3}}}{{2}},\frac{{{3}+{3}}}{{2}}\right)}={\left({0},{3}\right)}\)

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

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

\(\displaystyle{2}{a}={\left|{{3}-{\left(-{3}\right)}}\right|}\)

\(\displaystyle{2}{a}={\left|{{6}}\right|}={6}\)

\(\displaystyle{a}={3}\)

\(\displaystyle{a}^{2}={9}\)

The coordinates of the foci are \(\displaystyle{\left({h}\pm{c},{k}\right)}\).

So, we have \(\displaystyle{\left({h}-{c},{k}\right)}={\left(-{4},{3}\right)}{\quad\text{and}\quad}{\left({h}+{c},{k}\right)}={\left({4},{3}\right)}.\)

Solve the above coordinates for c as follows:

\(\displaystyle{\left({h}-{c},{k}\right)}={\left(-{4},{3}\right)}\)

\(\displaystyle{h}-{c}=-{4},{k}={3}\)

\(\displaystyle{0}-{c}=-{4}\)

\(\displaystyle{\left({h},{k}\right)}={\left({0},{3}\right)}\)

\(\displaystyle{c}={4}\)

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:

\(\displaystyle{c}^{2}={a}^{2}+{b}^{2}\)

\(\displaystyle{4}^{2}={3}^{2}+{b}^{2}\)

\(\displaystyle{b}^{2}={16}-{9}\)

\(\displaystyle{b}^{2}={7}\)

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

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

(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 \(\displaystyle{p}=-{4}+{2}=-{2}.\)

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

\(\displaystyle{\left({y}-{k}\right)}^{2}={4}{p}{\left({x}-{h}\right)}\)

\(\displaystyle{\left({y}-{4}\right)}^{2}={4}{\left(-{2}\right)}{\left({x}+{2}\right)}\)

\(\displaystyle{\left({y}-{4}\right)}^{2}=-{8}{\left({x}+{2}\right)}\)