I'm sure how to work the following problems cant anyone help?
1) Write an equation for the parabola with focus (4,0) and directrix y=2.
What do I need to do before I graph the following equations?
e1) 16x^2+9y^2=144
e2) x^2/9 - (y+2)^2/9 = 1
the focus is at y = 0 and the directrix is at y = 2 so the parabola opens down (sheds water.)
The distance from the focus to the vertex is the same as from the directrix to the vertex ( call it a ) so the vertex is at (4,1), halfway between directrix and focus
now so a = 1 but if parabola is upside down, use negative a so a = -1
with vertex at (h,k) which is (4,1)
form is
(x-h)^2 = 4 (a)(y-k)
(x-4)^2 = -4 (y-1)
x^2 - 8x + 16 = -4 y + 4
- 4 y = x^2 - 8 x + 16
y = -(x^2)/4 + 2 x - 4
e1) 16x^2+9y^2=144
this is an ellipse
(16/144)x^2 + (9/144) y^2 = 1
x^2/3^2 + y^2/4^2 = 1
center at (0,0)
x half axis 3
y half axis 4
e2) x^2/3^2 - (y+2)^2/3^2 = 1
hyperbola
center at (0, -2)
slopes of asymptotes = 3/3 = 1 (or -1)
vertex at (3,-2) and at (-3,-2)
For the first problem, to write an equation for a parabola with a given focus and directrix, you need to understand the definition of a parabola and its geometric properties.
1) Write an equation for the parabola with focus (4,0) and directrix y=2:
The standard form of the equation for a parabola in general is (x - h)^2 = 4p(y - k), where (h, k) is the vertex and p is the distance from the focus to the vertex and from the directrix to the vertex, with p being a positive value for a parabola that opens upward.
In this case, the given focus is (4,0), so the vertex is also at (4,0). The directrix is y=2, which means the distance from the vertex to the directrix is 2 units.
To find the value of p, calculate the distance between the focus and the vertex (or the directrix and the vertex). In this case, the distance between the focus (4,0) and the vertex (4,0) is 0 units. So, p = 0.
Plugging the values into the standard equation form, you get (x - 4)^2 = 4(0)(y - 0), which simplifies to (x - 4)^2 = 0.
Therefore, the equation for the given parabola is (x - 4)^2 = 0.
Now, let's move on to the second part of your question:
Before graphing the given equations, it's helpful to identify their key properties and any specific forms they may have.
e1) 16x^2 + 9y^2 = 144 is in the standard form of an ellipse equation: (x-h)^2/a^2 + (y-k)^2/b^2 = 1.
Here, the major axis is along the x-axis with the center at (h, k). The semi-major axis is "a," and the semi-minor axis is "b."
In this case, by comparing the given equation to the standard form, we can identify that the center is at (0,0).
The semi-major axis can be found as the square root of the value dividing the x-term, which is √(144/16) = 3.
The semi-minor axis can be found as the square root of the value dividing the y-term, which is √(144/9) = 4.
e2) x^2/9 - (y+2)^2/9 = 1 is in the standard form of a hyperbola equation: (x-h)^2/a^2 - (y-k)^2/b^2 = 1.
Here, the transverse axis is along the x-axis with the center at (h, k). The semi-transverse axis is "a," and the semi-conjugate axis is "b."
Comparing the given equation to the standard form, we can identify that the center is at (0,-2).
The semi-transverse axis can be found as the square root of the value dividing the x-term, which is √9 = 3.
The semi-conjugate axis can be found as the square root of the value dividing the y-term, which is √9 = 3.
By understanding the properties and forms of these equations, you can now proceed to graphing them using the identified properties and key points.