|
Question 39221: Identify the conic section represented by the equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center and the radius. For an ellipse or hyperbola, give the center and the foci. Sketch the graph.
16x^2+96x-9y^2+36y=36
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES AND TRY .IF STILL IN DIFFICULTY PLEASE COME BACK
------------------------------------------------------
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
1 solutions
Answer 22495 by venugopalramana(1898) About Me on
2006-05-07 08:31:50 (Show Source):
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
graph( 500, 500, -20, 20, -20, 20,(x^2-4*x-12)/8 )
Quadratic-relations-and-conic-sections/36550: 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.
9x^2 – 96y = 16y^2 + 18x + 279
1 solutions
Answer 22493 by venugopalramana(1898) About Me on
2006-05-07 08:27:57 (Show Source):
TIP
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS WITH
OPPOSITE SIGNS THEN IT
COULD BE HYPERBOLA
3.) 9x2 – 96y = 16y2 + 18x + 279
COMPLETE SQUARE...
{(3X)^2-2*3X*3+3^2}-3^2-{(4Y)^2+2*4Y*12+12^2}-12^2=279
(3X+3)^2-(4Y+12)^2=279+9+144=432
9(X+1)^2-16(Y+3)^2=432......NOW DIVIDE THROUGH OUT
WITH 432 TO GET 1 ON THE RHS.
(X+1)^2/(432/9) -(Y+3)^2/(432/16)=1
(X+1)^2/48 - (Y+3)^2/27 =1
THIS THE EQN. OF A HYPERBOLA.STD.EQN.IS.
(X-H)^2/A^2 - (Y-K)^2/B^2=1..WHERE
(H,K) IS CENTRE.....(-1,-3) HERE
TRANSVERSE AXIS IS Y=K...Y=-3
LENGTH OF TRANSVERSE AXIS=2A..
...=2SQRT(48)
CONJUGATE AXIS IS X=H.......X=-1
LENGTH OF CONJUGATE AXIS = 2B
=2SQRT(27)
ECCENTRICITY=E=SQRT{(A^2+B^2)/A^2}
=SQRT{(48+27)/48}=SQRT(75/48)
A*E=SQRT(48)*SQRT(75/48)=SQRT(75)
FOCI ARE (H+-AE,K)......(-1+SQRT(75),-3)
AND ......(-1-SQRT(75),-3)
A/E=SQRT(48)/SQRT(75/48)=48/SQRT(75)
DIRECTRIX ARE X=H+-A/E....
X=-1+48/SQRT(75)...AND
X=-1-48/SQRT(75)
ASYMPTOTES ARE GIVEN BY
(X-H)^2/A^2 = (Y-K)^2/B^2
OR
(X-H)/A=+(Y-K)/B AND............(X+1)/SQRT(48)
=(Y+3)/SQRT(27)
(X-H)/A=-(Y-K)/B.............(X+1)/SQRT(48) =
-(Y+3)/SQRT(27)
GRAPH IS GIVEN BELOW..
graph( 500, 500, -50, 50, -50, 50,
-3+27*(((x+1)^2-48)/48)^0.5,-3-27*(((x+1)^2-48)/48)^0.5)
--------------------------------------------------------
Quadratic-relations-and-conic-sections/36551: 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.
4(x-1)2 = 4-y^2
1 solutions
Answer 22486 by venugopalramana(1898) About Me on
2006-05-07 06:33:13 (Show Source):
TIP:
IF X^2 AND Y^2 HAVE DIFFERENT COEFFICIENTS OF SAME
SIGN ,THEN IT COULD
BE ELLIPSE.
4.) 4(x-1)2 = 4-y2
I HOPE IT IS
4(X-1)^2+Y^2=4...DIVIDE WITH 4
(X-1)^2/1^2+Y^2/2^2=1
THIS THE EQN.OF AN ELLIPSE.STD.EQN. IS
(X-H)^2/A^2 +(Y-K)^2/B^2=1
WHERE
(H,K) IS CENTRTE....(1,0).HERE
LENGTH OF MAJOR AXIS IS 2B...2*2=4
MINOR AXIS IS ALONG Y=K....Y=0
LENGTH OF MINOR AXIS IS 2A...2*1=2
ECCENTRICITY=E=SQRT{(B^2-A^2)/A^2}
=SQRT(4-1)/1=SQRT(3)
B*E=2SQRT(3)
B/E=2/SQRT(3)
FOCI ARE (H,K+BE) AND (H,K-BE)
(1,2SQRT(3)) AND (1,-2SQRT(3))
DIRECTRIX ARE.Y=K+B/E AND K-B/E
Y=2/SQRT(3)..AND...Y=-2/SQRT(3)
GRAPH IS GIVEN BELOW...
graph( 500, 500, -3, 3, -3, 3,
(4-4*(x-1)^2)^0.5,-(4-4*(x-1)^2)^0.5)
|
|
|
| |
