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Show that the shortest length of the nor...

Show that the shortest length of the normal chord to the parabola `y^(2) = `4ax is `6sqrt(3a)`

Answer

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Knowledge Check

  • The locus of poles of normal chords of the parabola y^(2)=4ax is

    A
    `(x+2a)y^(2)=4a^(3)`
    B
    `(x+2a)y^(2)+4a^(3)=0`
    C
    `(x-2a)y^(2)=4a^(3)`
    D
    `(x-2a)y^(2)+4a^(3)=0`
  • The locus of middle points of normal chords of the parabola y^(2) = 4ax is

    A
    ` (y^(2) )/(2a) + ( 4a^(3))/( y^(2))= x-2a`
    B
    ` (y^(2))/(2a) -(4a^(3))/(y^(2))=x-2a`
    C
    ` (y^(2))/( 2a) +(4a^(3))/( y^(2))=x+2a `
    D
    ` (y^(2))/(2a) -(4a^(3))/( y^(2))=x-2a `
  • The subnormal of the parabola y^(2)=4ax is equal to

    A
    latus rectum
    B
    semi latus rectum
    C
    2(latus rectum )
    D
    none
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