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‘O’ is the vertex of the parabola y^(2) ...

‘O’ is the vertex of the parabola `y^(2) = 4ax` & L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H, prove that the length of the double ordinate through H is `4asqrt5` .

Answer

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'O' is the vertex of the parabola y^(2)=4ax&L is the upper end of the latus rectum.If LH is drawn perpendicular to OL meeting OX in H, prove that the length of the double ordinate through H is 4a sqrt(5).

The normal to y^(2)=4a(x-a) at the upper end of the latus rectum is

Knowledge Check

  • The equation of tangents to the parabola y^2 = 4ax at the ends of latus rectum is :

    A
    `x-y + a=0`
    B
    `x+y+a=0`
    C
    `x+y-a=0`
    D
    both (a) and (b)

  • If the probola y^(2) = 4ax passes through (-3, 2), then length of its latus rectum is

    A
    `2/3`
    B
    `1/3`
    C
    `4/3`
    D
    `4`

  • If ASC is a focal chord of the parabola y^(2)=4ax and AS=5,SC=9 , then length of latus rectum of the parabola equals

    A
    `(90)/(7)`
    B
    `(7)/(90)`
    C
    `(45)/(14)`
    D
    `(14)/(45)`

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    If the line 2x+3y=1 touches the parabola y^(2)=4ax then the length of latus rectum is

    In the parabola y^2 = 4x , the ends of the double ordinate through the focus are P and Q. Let O be the vertex. Then the length of the double ordinate PQ is

    Let p be a point on the parabola y^(2)=4ax then the abscissa of p ,the ordinates of p and the latus rectum are in

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