Question

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

Conic sections
ANSWERED
asked 2021-01-06
Write the equation of each conic section, given the following characteristics:
a) Write the equation of an ellipse with center at (3, 2) and horizontal major axis with length 8. The minor axis is 6 units long.
b) Write the equation of a hyperbola with vertices at (3, 3) and (-3, 3). The foci are located at (4, 3) and (-4, 3).
c) Write the equation of a parabola with vertex at (-2, 4) and focus at (-4, 4)

Answers (1)

2021-01-07
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:
\(\displaystyle{\left({y}-{k}\right)}^{2}={4}{p}{\left({x}-{h}\right)}\)
\(\displaystyle{\left({y}-{4}\right)}^{2}={4}{\left(-{2}\right)}{\left({x}+{2}\right)}\)
\(\displaystyle{\left({y}-{4}\right)}^{2}=-{8}{\left({x}+{2}\right)}\)
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