To solve this problem you must apply the proccedure shown below:
1. You have that the ellipse given as a vertical major axis (a=13), therefore, taking the ellipse with its center at the origin, you have the following equation:
(y^2/a^2)+(x^2/b^2)=1
2. You have the distance from the center of the ellipse to the focus:
c=12, therefore, you can calculate the value of b, the minor radius:
c^2=a^2-b^2
b=√(13^3-12^2)
b=5
3. Therefore, the equation is:
a^2=169
b^2=25
(y^2/169)+(x^2/25)=1
The answer is: (y^2/169)+(x^2/25)=1
All the triangles are isosceles meaning that they have to sides that are equal
The graph graphed is y=1/2x
so
y=1/2x and y=-x+6
y=y so
1/2x=-x+6
times 2 both sides
x=-2x+12
add 2x both sides
3x=12
x=4
sub back
y=1/2x
y=1/2(4)
y=2
(4,2) is the solution
Make a line of best fit then find the slope and y intercept. put it into the y=mx + b form.
m=slope
b=y intercept
Let the equation be:
y = ax^2 + bx + c.
Then, substitue the three points into the equation.
First point: 0 = a0^2 + b0 + c.
So c = 0.
Second point: -2 = a(-1)^2 + b(-1) + c.
So a - b + c = -2.
Third point: 6 = a*1^2 + b*1 + c.
So a + b + c = 6.
We know that c=0 already, so we substitute c=0 into the last two equations and we would get:
a - b = -2
a + b = 6
We add the two equations and we get:
2a = 4
a = 2
Then, we substitute a=2 into a-b=-2 and we get:
-b = -4
b = 4
Now we know a = 2, b = 4, and c = 0
Then, the equation of the parabola would be:
2x^2 + 4x
