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The line x/a+y/b=1 moves in such a way t...

The line `x/a+y/b=1` moves in such a way that `1/(a^(2))+1/(b^(2))=1/(c^(2))` where ic si a constnat. The locus of the foot of perpendicular from the origin on the given line is `x^(2)+y^(2)=c^(2)`

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If the line ((x)/(a))+((y)/(b))=1 moves in such a way that ((1)/(a^(2)))+((1)/(b^(2)))=((1)/(c^(2))), where c is a constant,prove that the foot of the perpendicular from the origin on the straight line describes the circle x^(2)+y^(2)=c^(2)

The locus of the foot of the perpendicular from the origin on each member of the family (4a+3)x-(a+1)y-(2a+1)=0

Knowledge Check

  • The locus of the foot of the perpendicular from the foci an any tangent to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , is

    A
    `x^(2) + y^(2) = b^(2)`
    B
    `x^(2) + y^(2) = a^(2)`
    C
    `x^(2) + y^(2) = a^(2) + b^(2)`
    D
    none of these

  • For a variabkle line x/a+y/b=1 where 1/(a^(2))+1/(b^(2))=1/(c^(2)) the locus of the foot of perpendicular drawn from origin to it is

    A
    `x^(2)+y^(2)=c^(2)//2`
    B
    `x^(2)+y^(2)=c^(2)`
    C
    `x^(2)+y^(2)=2c`
    D
    None of these

  • The foot of perpendicular from the point (1,2,3) to the line (x)/(2)=(y-1)/(3)=(z-1)/(3) is

    A
    `(1,(5)/(2),(5)/(2))`
    B
    `(1,(9)/(4),(11)/(4))`
    C
    `(1,3,2)`
    D
    `(3,1,2)`

    • Similar Questions

    Explore conceptually related problems

    Find the locus of the foot of perpendicular from the centre upon any normal to line hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 .

    if (x)/(a)+(y)/(b)=1 is a variable line where (1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))(c is constant ) then the locus of foot of the perpendicular drawn from origin

    Find the coordinates of the foot of the perpendicular drawn from the point (1,-2) on the line y=2x+1

    If 2p is the length of perpendicular from the origin to the lines (x)/(a)+(y)/(b)=1, then a^(2),8p^(2),b^(2) are in

    The foot of perpendicular from (0,2,3) to the line (x+3)/(5)=(y-1)/(2)=(z+4)/(3) is

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