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Question 35839This question is from textbook Precalculus: A Graphing Approach
: Hi! I'm working on some Conic Section problems, and do ok when they are centered at zero. My problem is when their vertex has shifted and I am asked to graph a problem with a given equation or if I'm given the foci/directrix and am required to write the equation. If you could clear this up for me that would be wonderful!
Write the standard equation:
y^2-4x+4=0
Find the standard form of the equation of the parabola with focus (8,-2) and directrix x=4
Thanks so much for taking time to look at my question!
~Troubled Conic Student :)
This question is from textbook Precalculus: A Graphing Approach
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Hi! I'm working on some Conic Section problems, and do ok when they are centered at zero. My problem is when their vertex has shifted and I am asked to graph a problem with a given equation or if I'm given the foci/directrix and am required to write the equation. If you could clear this up for me that would be wonderful!
Write the standard equation:
y^2-4x+4=0
STANDARD FORM OF PARABOLA IS
(Y-K)^2=4A(X-H),WHERE (H,K) IS VERTEX,FOCUS IS (H+A,K),DIRECTRIX IS X-H+A=0,
AND AXIS IS Y-K=0.HENCE WE GET HERE AS STANDARD FORM
(Y-0)^2=4(X-1)....H=1...K=0...A=1...
VERTEX = (1,0)
FOCUS IS (1+1,0)=(2,0)
DIRECTRIX IS X-1+1=0....OR....X=0
AXIS IS Y-0=0...Y=0
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Find the standard form of the equation of the parabola with focus (8,-2) and directrix x=4
FOCUS .....H+A=8......I................K=-2
DIRECTRIX=X-H+A=0...OR....X=H-A=4............II
EQN.I+EQN.II GIVES
H+A+H-A=8+4=12
2H=12
H=6
A=8-H=2
HENCE EQN.IS
(Y-K)^2=4A(X-H)
(Y+2)^2=4*2*(X-6)=8(X-6)
Thanks so much for taking time to look at my question!
~Troubled Conic Student :)
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