Finding the general and standard form equations for the circle and parabola passing through thee points (4,6), (-6,2), and (1,-3).
I used the y=Ax^2+Bx+C to find a,b,c.
(4,6)=> 6=A(4)^2+B(4)+C
(-6,2)=> 2=A(-6)^2+B(-6)+C
(1,-3)=> -3A(1)^2+B(1)+C
then,
a=13/15, b=8/7, and c=-158/35
so the equation for parabola:
how do I make this equation into this equation f(x)a(x-h)^2+k without the vertex?
I tried to find the equation for the circle and parabola from this three points (4,6),(-6,2), and (1,-3). this is what i get:
for parabola:
general form=
standard form=
Vertex: (158,8x/7)
focus: (58, 8x/7, -35/4)
directrix: y=8x/7 -35/4
for circle:
radius = 13.4
center= (77/6, 85/22)
Am I right? please help me out... thanks!
The general equation of the circle passing through (4, 6), (-6, 2) and (1, -3) is
Substitute the points in the above equation, and solve for g, f and c. I got the equation of the circle as
Check it.
Consider any one point as the common point for circle and parabola.
Let it be the vertex. Find the equation of the tangent to the circle at the vertex in the form ax +by + c = 0. Diretrix of the parabola will be parallel to this tangent in the form ax + by + c' = 0.
Let (h, k) be the focus. In the parabola, the distance between any point and the focus is equal the distance between the point and diretrix.
Write down three equations for three points and solve for h, k and c'
Now we need to find the circle: (x - h)^2 + (y - k)^2 = r^2
center = (h,k)
radius = r
So we solve another system of 3 equations:
(4 - h)^2 + (6 - k)^2 = r^2
(-6 - h)^2 + (2 - k)^2 = r^2
(1 - h)^2 + (-3 - k)^2 = r^2
[(4 - h)^2 + (6 - k)^2 = r^2] - [(-6 - h)^2 + (2 - k)^2 = r^2] = (-20h - 8k + 12 = 0)
[(4 - h)^2 + (6 - k)^2 = r^2] - [(1 - h)^2 + (-3 - k)^2 = r^2] = (-6h - 18k + 42 = 0)
Now solve the system of two equations:
(-20h - 8k + 12 = 0)
(-6h - 18k + 42 = 0)
h = (-5/13)
k = (32/13)
r = sqrt(5365)/13
x + (5/13)]^2 + [y - (32/13)]^2 = [sqrt(5365)/13]^2
this is what i got.......