Quadratic Relations and Conic Sections Unit
Does any have pictures or the steps on how to do them? I am falling plz thanks!!!!!
1. A searchlight has a parabolic reflector that forms a "bowl," which is 18 inches wide from rim to rim and 10 inches deep. If the filament of the lightbulb is located at the focus, how far from the vertex of the reflector is it? Round your answer to the nearest hundredth. You may want to draw a graph to help you with this problem. Include it as part of your answer.
2. Find the equation of a circle with center at (-6, 4) and radius length 50. Then, give the domain and range of the relation. Give the answer in the simplest radical form, NOT IN DECIMAL FORM.
3. Find the equation of a parabola given the following information:
Focus (5, 3) and directrix: x = -5
4. Write the equation of the ellipse with foci at (0, ±24) and minor axis length 14.
5) Write the equation of this hyperbola. The asymptotes are included in green to help you. (Hint: You'll need to figure out the equations of the asymptotes to figure this one out). Then, give the equation of this same hyperbola translated 3 units right and 2 units down.
I love it when people post wrong answers!! - the teacher of this course!
Yes, you are failing this course. - your teacher.
LMFAOOO THIS IS TOO FUNNY
#1. You can model this parabola as
y = ax^2 - 10
Since y(9) = 0, that gives you
81a-10 = 0
a = 10/81
y = 10/81 x^2 - 10
Now, rewrite that as
x^2 = 81/10 (y+10)
Now recall that x^2 = 4py as its vertex p units from the focus.
So 4p = 81/10
p = 81/40
That is the distance from the vertex to the focus
#2. (-6, 4) and radius length 50
(x+6)^2 + (y-4)^2 = 50^2
domain: [-6-50,-6+50]
range: [4-50,4+50]
#3. Focus (5, 3) and directrix: x = -5
recall that y^2 = 4px has
focus at (p,0) and directrix x = -p
The vertex is midway between the focus and the directrix, at (0,3)
so the equation is
(y-3)^2 = 20x
#4. ellipse with foci (0, ±24) and minor axis length 14.
so now we have a vertical major axis, with
b = 7
c = 24
a^2 = b^2+c^2 = 25
That gives the equation
x^2/49 + y^2/576 = 1
#5. No graphs, but it will be of the form
(x-3)^2/a^2 - (y+2)^2/b^2 = 1
if the axis is horizontal.
Better review the properties of hyperbolas.