Question 21167: sorry,the 1st letter had some mistakes. hi.I have a question how can you find the coordinates of the vertex.Make a table of values,using x-values to the left and to the right of the vertex.
1.y=3x^2
how can we now "b" and "c" with this kind of equation.there`s no graphs sktechings.
Thank you.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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(545) 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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