Question 936146: find the equation of the parabola with axis of symmetry parallel to x-axis,latus rectum 1 and passing through (3,1),(-5,5)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the equation of the parabola with axis of symmetry parallel to x-axis,latus rectum 1 and passing through (3,1),(-5,5)
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Basic equation of parabola with axis of symmetry parallel to x-axis:
(y-k)^2=±4p(x-h),(h,k)=coordinates of the vertex
latus rectum=1=4p
(y-k)^2=(x-h)
..
set up 2 equations:
(3,1) (1-k)^2=(3-h)
(-5,5)(5-k)^2=(-5-h)
expand equations:
1-2k+k^2=3-h
25-10k+k^2=-5-h
subtract:
-24+8k=8
8k=32
k=4
..
1-8+16=3-h
h=-6
..
(h,k)=(-6,4)
Equation: (y-4)^2=(x+6)
This is a parabola that opens rightward with vertex at (-6,4)
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