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Question 35337: 1. graph x-3= -1/8(y+2)^2. Write the coordinates of the vertex and the focus and the equation of the directrix.
2.Find all solution to each system of equations algerbaiclly.
x^2 + y^2 =25
3x^2 + 2y^2 = 59
3. Write the standard equations of the conic section with the given characteristics
hyperbola with foci at (5,-7) and (14,0) and vertices at (12,0) and (4,0)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 1. graph x-3= -I/8 (y=2)^2. Write the coordinates of the vertex and
> the
> > > focus and the equation of the directrix.
X-3=-(1/8)(Y-2)^2
(Y-2)^2=-8(X-3)
COMPARING WITH STD.EQN.OF PARABOLA
(Y-K)^2=4A(X-H)^2...WE GET
VERTEX=(H,K)=(3,2)
A=-2
FOCUS =(H+A,K)=(3-2,2)=(1,2)
DIRECTIX IS ... X-H+A=0...THAT IS
X-3-2=0...OR...X=5
GRAPH IS GIVEN BELOW
> > > 2.Find all solution to each system of equations algerbaiclly.
> > > x^2 + y^2 =25
SINCE SUM OF TWO SQUARES = 25 ,AND SQUARES ARE ALWYS POSITIVE ,WE GET THE SOLUTION SETS AS X^2<=25 AND Y^2<=25....THAT IS
1.0<=|X|<=5 AND 0<=|Y|<=5
> > > 3x^2 + 2y^2 = 59...YOU CAN DO THE SAME WAY AS ABOVE
3X^2<=59 AND 2Y^2<=59..ETC.......
> > > 3. Write the standard equations of the conic section with the given
> > > characteristics
> > > hyperbola with foci at (5,-7) and (14,0) and vertices at (12,0) and
> > > (4,0)
STANDARD EQN.OF HYPERBOLA IS
(X-H)^2/A^2 -(Y-K)^2/B^2 =1...WHERE
CENTRE IS (H,K)...WE ARE GIVEN VERTICES ARE AT (12,0) AND (4,0)...SO CENTRE IS AT
(12+4)/0,(0+0)/2=(8,0)
HENCE H=8....K=0
TRANSVERSE AXIS IS Y=K=0
LENGTH OF TRANVERSE AXIS =2A=16
A=8
CONJUGATE AXIS IS X=H=8
PLEASE CHECK DATA....CENTRE WHICH IS MID POINT OF FOCI AND ALSO MIDPOINT OF VERTICES ARE NOT TALLYING
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