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

jernplate8

Answered question

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)

Answer & Explanation

Anonym

Anonym

Skilled2021-01-07Added 108 answers

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.
(xh)2a2+(yk)2b2=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 2a=8,a=4. 
Since the length of the minor axis is 6, then 2b=6,b=3. 
Consequently, the ellipse's basic equation (3, 2) and a=4,b=3 is given below: 
(x3)242+(y2)232=1 
(x3)216+(y2)29=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 (xh)2a2(yk)2b2=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:
(h,k)=(332,3+32)=(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:
2a=|3(3)| 
2a=|6|=6 
a=3 
a2=9 
The coordinates of the foci are (h±c,k)
So, we have (hc,k)=(4,3) and (h+c,k)=(4,3). 
The aforementioned coordinates for c should be solved as follows:
(hc,k)=(4,3) 
hc=4,k=3 
0c=4 
(h,k)=(0,3) 
c=4 
Now obtain the value of b2 
using the relation c2=a2+b2 with a=3,c=4 as follows: 
c2=a2+b2 
42=32+b2 
b2=169 
b2=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:
(x0)29(y3)27=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 p=4+2=2. 
As a result, the following equation for the parabola with vertex (-2, 4) and focus (-4, 4) is provided:
(yk)2=4p(xh) 
 

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