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{:(,"Hyperbola point ",,"Centre"),(I,4(x...

` {:(,"Hyperbola point ",,"Centre"),(I,4(x+3) ^(2) -9(y-2)^(2) =36",", ,(a) ("2,-1")),(II,3(x-3) ^(2) -4(y-1)^(2) =12",",,(b) ("-3,2")),(III,9(x-2) ^(2) (2) -5(y-1) ^(2) = 45"," ,,(c) (2,1)) ,(IV,x^(2) -4x-y^(2) -2y -8=0",",,(d) ("3,1")):}`

A

a,b,c,d

B

b,d,c,a

C

c,a,b,d

D

b,c,d,a

Text Solution

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The correct Answer is:
B
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Match the following . {:("Circles"," Radical centre"),(I.x^(2) + y^(2) = 1" ," x^(2) + y^(2) - 2x = 1 "," x^(2) + y^(2) - 2y = 1 , " a)(0,0)"),(II. x^(2) + y^(2) - x + 3y - 3 = 0 ", " x^(2) + y^(2) - 2x + 2y + 2 = 0 ", " x^(2) + y^(2) + 2x + 3y - 9 = 0, "b)(2,3)"),(III.x^(2) + y^(2) - 8x + 40 = 0 "," x^(2) + y^(2) - 5x + 16 = 0"," x^(2) + y^(2) - 8x + 16y + 160 = 0 ,"c)(8,-15/2)") :}

To remove the first degree terms in the following equations origin should be shifted to the another point then calculate the new origins from list - II {:(" List - I "," List - II "),("(A) "x^(2)-y^(2)+2x+4y=0,"(1) (5,-7) "),("(B) "4x^(2) +9y^(2)-8x+36y + 4 = 0,"(2) (1,-2) "),("(C) "x^(2) + 3y^(2) + 2x + 12y + 1 = 0,"(3) (-1,2) "),("(D) "2(x-5)^(2)+3(y+7)^(2)=10,"(4) (-1,-2) "),(,"(5) (-5,7) "):} The correct matching is

Knowledge Check

  • {:(,"Hyperbola point ",,"Pole"),(I,3x^(2)-4y^(2) =12-3x+2y+6=0",", ,(a) ("2,-3")),(II,x^(2) -3y^(2) =32x + y-1=0",",,(b) ("-1,-3")),(III,3x^(2) -5y^(2) =15-2x+5y-5=0"," ,,(c) (-2,1)) ,(IV,9x^(2) -16y^(2) =144-3x-16y+48=0",",,(d) ("6,-1")):}

    A
    a,b,c,d
    B
    a,b,d,c
    C
    c,d,a,b
    D
    b,c,d,a

  • {:(,"Hyperbola point ",,"Tangent"),(I,3x^(2) -4y ^(2) =12","3x+2y +6=0, ,(a) ("2,-3")),(II,x^(2) -3y^(2) =3","2x+ y-1 =0,,(b) ("-1,-3")),(III,3x^(2) -4y^(2) =8","(2","-1) ,,(c) x+2y-4=0),(IV,9x^(2) -16y^(2) =144","3x-16y + 48 =0,,(d) ("6,-1")):}

    A
    a,b,c
    B
    b,c,a
    C
    c,b,a
    D
    a,c,b

  • {:(,"Hyperbola",,"Foci "),(I,((x-1) ^(2) )/(16) - ((y-2)^(2) )/(9) =1,,(a) ("1,-1")("-9,-1")),(II,((x+2)^(2) )/(9) - ((y-3) ^(2) )/(27) =1 ,,(b) ("6,2") ("-4,2")),(III,((x+1)^(2) )/(25) -((y+2)^(2) )/(16)= 1 ,,(c) ("4,3") ("-8,3")),(IV,9x^(2) -4y(2) =8","(2","-) ,,(d) (-1pm sqrt (41) ","-2)):}

    A
    a,b,c,d
    B
    a,d,b,c
    C
    c,a,b,d
    D
    b,c,d,a

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    (x+3)/(2)-y=2,(x-3)/(2)+2y-4(1)/(2) then x=……..

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