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The normal at a point t1 on y^2 = 4ax m...

The normal at a point `t_1` on `y^2 = 4ax` meets the parabola again in the point `t_2`. Then prove that `t_1 t_2 + t_1^2 + 2=0`

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Knowledge Check

  • The normal at a point (bt_1^(2) , 2bt_1) on a parabola meets the parabola again in the point (bt_2^(2) , 2bt_2) then

    A
    ` t_2 =-t_1 +(2)/( t_1) `
    B
    ` t_2 = t_1 +(2)/( t_1) `
    C
    ` t_2 =t_1 +(2)/(t_1) `
    D
    ` t_2=-t_1- (2)/(t_1) `

  • If the normals at t_(1) and t_(2) on y^(2) = 4ax meet again on the parabola then t_(1)t_(2) =

    A
    1
    B
    `-1`
    C
    2
    D
    `-2`

  • If the normal to y^(2) =4ax at t_1 cuts the parabola again at t_2 then

    A
    ` t_2^(2) le 8`
    B
    ` t_2 ^(2) ge 8`
    C
    ` -8 le t_2 le 8`
    D
    ` t_2^(2) lt 8`

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    If the normals at the points t_(1) and t_(2) on y^(2) = 4ax at the point t_(3) on the parabola, the t_(1)t_(2) =

    If the normals at the points t_(1) and t_(2) on y^(2) = 4ax at the point t_(3) on the parabola, the t_(1)t_(2) =

    If the normals at the point t_(1) and t_(2) on y^(2)=4ax intersect at the point t_(3) on the parabola then t_(1)t_(2) =

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