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Question 22088: classify conic section and write equation in standard form x^2+4y^-2x-3=0, 2x^2=z0x-y=41=0, 5x62-3y^2-30=0, x^2=y^2-12x=4y=31=0, y^2-8x-4y=4=0, -x^2+y^2-6x-6y-4=0
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES TO UNDERSTAND THE TOPIC AND TRY YOUR PROBLEMS
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classify the conic section and write each equation in standard form.
x^2-8x+4y+16=0, 3x^2+3y^2-30x+59=0, x^2+2y^2-8x+7=0, 4x^2-y^2+16x+6y-3=0, 3x^2=y^2-4y+3=0, x^2+y^2-2x+1oy+1=0
1 solutions
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Answer 11684 by venugopalramana(594) on 2005-12-27 09:53:02 (Show Source):
classify the conic section and write each equation in standard form.
YOU HAVE TO WRITE THEM SO THAT ALL SECOND DEGREE TERMS ,TAKING THE FIRST DEGREE TERMS ALSO AS AVAILABLE, ARE PUT IN AS COMBINATION OF PERFECT SQUARES...THEN YOU MAY END UP AS FOLLOWS..IN GENERAL ,THERE ARE FEW OTHER CRITERIA TO BE FOLLOWED ,BUT WE WILL LEAVE THEM FOR THE PRESENT....
1.ALL SECOND DEGREE TERMS COMBINE TO FORM A PERFECT SQUARE..SOME FIRST DEGREE TERMS AND CONSTANT MAY BE LEFT OUT..THEN IT IS A PARABOLA.(PROBLEM 1 BELOW)
2.SECOND DEGREE TERMS OF X COMBINE TO FORM A PERFECT SQUARE AND SECOND DEGREE TERMS OF Y COMBINE TO FORM A PERFECT SQUARE.THEN IT IS AN ELLIPSE OR HYPERBOLA
BASED ON THE FOLLOWING DIFFERENCE
2A.IF THE X AN Y TERMS AS COMBINED INTO PERFECT SQUARES ARE HAVING SAME SIGN THEN IT IS AN ELLIPSE.(PROBLEM 2 BELOW)
2B.IF THE X AN Y TERMS AS COMBINED INTO PERFECT SQUARES ARE HAVING OPPOSITE SIGN THEN IT IS A HYPERBOLA.(PROBLEM 4 BELOW)
1.x^2-8x+4y+16=0
{X^2-2*X*4+4^2}-4^2+4Y+16=0
(X-4)^2+4Y=0
(X-4)^2=4*(-1)Y........THIS IS IN STANDARD FORM
(X-H)^2=4*A*(Y-K) OF PARABOLA...SO THIS IS A PARABOLA
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2.3x^2+3y^2-30x+59=0
{3(X^2-10X)}+3Y^2+59=0
[3{(X^2-2*X*5+5^2)-5^2}]+3Y^2+59=0
[3{(X-5)^2-25}]+3Y^2+59=0
3(X-5)^2-75+3Y^2+59=0
3(X-5)^2+3Y^2-16=0
3(X-5)^2+3Y^2=16.......DIVIDING WITH 16 THROUGH OUT,WE GET..........
(X-5)^2/(16/3)+Y^2/(16/3)=1........THIS IS IN THE STANDARD FORM
(X-H)^2/(A^2)+(Y-K)^2/(B^2)=1.....OF ELLIPSE.SO THIS IS AN ELLIPSE
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3.x^2+2y^2-8x+7=0
TRY YOUR SELF AS ABOVE
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4.4x^2-y^2+16x+6y-3=0
4(X^2+4X)-(Y^2-6Y)-3=0
4{(X^2+2*X*2+2^2)-2^2}-{(Y^2-2*Y*3+3^2)-3^2}-3=0
4(X+2)^2-(Y-3)^2-16+9-3=0
4(x+2)^2-(y-3)^2=10.....DIVIDING THROUGHOUT WITH 10
(X+2)^2/(10/4)-(Y-3)^2/10=1.....THIS IS IN THE STANDARD FORM
(X-H)^2/(A^2)-(Y-K)^2/(B^2)=1....OF HYPERBOLA.SO THIS IS A HYPERBOLA
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5.3x^2=y^2-4y+3=0
TRY YOUR SELF AS ABOVE
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6.x^2+y^2-2x+10y+1=0
TRY YOUR SELF AS ABOVE
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Find the foci of the given ellipse.
25x2+9y2+150x-90y+225=0
x2=xsquared(25xsquared)
y2=ysquared(ysquared)
1 solutions
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Answer 11617 by venugopalramana(594) on 2005-12-24 07:10:18 (Show Source):
25x2+9y2+150x-90y+225=0
{(5X)^2+2*5X*15+15^2}-15^2+{(3Y)^2-2*3Y*15+15^2}-15^2+225=0
(5X+15)^2+(3Y-15)^2=225
25(X+3)^2+9(Y-5)^2=225.....DIVIDING THROUGHOUT WITH 225
(X+3)^2/(225/25)+(Y-5)^2/(225/9)=1...COMPARING WITH STANDARD EQUATION OF ELLIPSE
((X-H)^2)/A^2)+((Y-K)^2)/B^2)=1,WE GET
A=SQRT(225/25)=15/9=5/3....B=SQRT(225/9)=15/3=5...
CENTRE IS (H,K)=(-3,5)
AND ECCENTRICITY = SQRT{(B^2-A^2)/B^2)}=SQRT((5^2-(5/3)^2)/5^2)
=SQRT(8/9).........
HENCE FOCI ARE (H,K+B*E) AND (H,K-B*E)....THAT IS
{-3,5+5SQRT(8/9)}AND {-3,5-5SQRT(8/9)}
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Please help me solve this equation: . I can not find the vertex, focus, or the directrix.
1 solutions
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Answer 11634 by venugopalramana(594) on 2005-12-25 11:41:25 (Show Source):
(X-2)^2=Y
THIS IS OF THE FORM (X-H)^2=4A(Y-K)..WHERE
H=2,K=0,A=1/4,.....
..HENCE THIS IS A PARABOLA WITH
FOCUS AS (H,K+A)..THAT IS (2,1/4)
VERTEX AS (H,K)...THAT IS (2,0)
AXIS AS X-H=0...THAT IS X-2=0
AND
DIRECTRIX AS Y-K+A=0...THAT IS Y+1/4=0
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