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The locus of the foot of perpendicular d...

The locus of the foot of perpendicular drawn from the centre of the ellipse `x^2+""3y^2=""6` on any tangent to it is (1) `(x^2-y^2)^2=""6x^2+""2y^2` (2) `(x^2-y^2)^2=""6x^2-2y^2` (3) `(x^2+y^2)^2=""6x^2+""2y^2` (4) `(x^2+y^2)^2=""6x^2-2y^2`

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Knowledge Check

  • The locus of the foot of prependicular drawn from the center of the ellipse x^(2)+3y^(2)=6 on any tangent to it is

    A
    `(x^(2)+y^(2))^(2)=6x^(2)+2y^(2)`
    B
    `(x^(2)+y^(2))^(2)=6x^(2)-2y^(2)`
    C
    `(x^(2)-y^(2))^(2)=6x^(2)+2y^(2)`
    D
    `(x^(2)-y^(2))^(2)=6x^(2)-2y^(2)`

  • The locus of the foot of perpendicular drawn from the centre of ellipse x^(2)+3y^(2)=6 on any tangent

    A
    `(x^(2)-y^(2))^2=6x^(2)+2y^(2)`
    B
    `(x^(2)-y^(2))^2=6x^(2)-2y^(2)`
    C
    `(x^(2)+y^(2))^2=6x^(2)+2y^(2)`
    D
    `(x^(2)+y^(2))^2=6x^(2)-2y^(2)`

  • The locus of feet of perpendicular from either foci of the ellipse (x – y +1)^(2) + (2x + 2y – 6)^(2) = 20 on any tangent will be :

    A
    `x^(2) + y^(2) + 2x + 4y + 5 = 0`
    B
    `x^(2) + y^(2) + 2x + 4y - 5 = 0`
    C
    `x^(2) + y^(2) - 2x - 4y - 5 = 0`
    D
    `x^(2) + y^(2) - 2x - 4y + 5 = 0`

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    Locus of the point of intersection of perpendicular tangents on the curve with centre (3/2, 7/2) and radius sqrt(15)/2 is: - (a) 2x^2+2y^2-4x-7y-1=0 (b) 2x^2+2y^2-14 x-7y-1=0 (c) 2x^2+2y^2-6x-14 y-1=0 (d) 2x^2-12 y^2-6x+14 y-1=0

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