 # 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 jernplate8 2021-01-06 Answered
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)
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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:
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1:a>b$
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 $2a=8,⇒a=4.$
Since the length of the minor axis is 6, then $2b=6,⇒b=3.$
Hence, the standard equation of the ellipse with center (3, 2) and $a=4,b=3$ is given below:
$\frac{{\left(x-3\right)}^{2}}{{4}^{2}}+\frac{{\left(y-2\right)}^{2}}{{3}^{2}}=1$
$\frac{{\left(x-3\right)}^{2}}{16}+\frac{{\left(y-2\right)}^{2}}{9}=1$
(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 $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1.$
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:
$\left(h,k\right)=\left(\frac{3-3}{2},\frac{3+3}{2}\right)=\left(0,3\right)$
The length of the transverse axis 2a, is bounded by the vertices.
Hence, the distance between x-coordinates is obtain as follows:
$2a=|3-\left(-3\right)|$
$2a=|6|=6$
$a=3$
${a}^{2}=9$
The coordinates of the foci are $\left(h±c,k\right)$.
So, we have $\left(h-c,k\right)=\left(-4,3\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(h+c,k\right)=\left(4,3\right).$
Solve the above coordinates for c as follows:
$\left(h-c,k\right)=\left(-4,3\right)$
$h-c=-4,k=3$
$0-c=-4$
$\left(h,k\right)=\left(0,3\right)$
$c=4$
Now obtain the value of ${b}^{2}$
using the relation 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:
$\frac{{\left(x-0\right)}^{2}}{9}-\frac{{\left(y-3\right)}^{2}}{7}=1$
(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 $p=-4+2=-2.$
Hence, the equation of the parabola with vertex (-2, 4) and focus (-4, 4) is given below:
${\left(y-k\right)}^{2}=4p\left(x-h\right)$

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