Question 36549: I was wondering if anyone could help me determine what type of figure is given, then graph. Show all work and label all key points (asymptotes, foci, vertices, the directrix, the center) where applicable.
x^2 – 4x – 8y = 12
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! TIP:
2.IF X^2 OR Y^2 IS ONLY PRESENT ,IT COULD BE A PARABOLA.
2.) x2 – 4x – 8y = 12
COMPLETE SQUARE
(X^2-2*2X+2^2)-2^2=12+8Y
(X-2)^2=8Y+16=8(Y+2)
THIS IS THE EQN.OF A PARABOLA .STD EQN. IS
(X-H)^2=4A(Y-K),WHERE
(H,K) IS THE VERTEX....(2,-2) HERE.
4A=LATUS RECTUM =8 HERE...A=2
FOCUS IS (H+A,K).....(2+2,-2)=(4,-2)..HERE.
DIRECTRIX IS X-H+A=0...
X-2+2=0...OR....X=0..
AXIS IS Y-K =0..Y+2=0
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