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If the lines joining the origin and the point of intersection of curves `a x^2+2h x y+b y^2+2gx+0` and `a_1x^2+2h_1x y+b_1y^2+2g_1x=0` are mutually perpendicular, then prove that `g(a_1+b_1)=g_1(a+b)dot`

Answer

Step by step text solution for If the lines joining the origin and the point of intersection of curves a x^2+2h x y+b y^2+2gx+0 and a_1x^2+2h_1x y+b_1y^2+2g_1x=0 are mutually perpendicular, then prove that g(a_1+b_1)=g_1(a+b)dot by MATHS experts to help you in doubts & scoring excellent marks in Class 11 exams.

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Knowledge Check

  • The lines joining the origin to the points of intersection of the curve ax^(2)+2hxy+by^(2)+2gx=0 and a_(1)x^(2)+2b_(1)xy+b_(1)y^(2)+2g_(1)x=0 are _|_ then

    A
    `(a+b)/(g_(1))=(a_(1)+b_(1))/g`
    B
    `(a+b)g_(1)=(a_(1)+b_(1))g`
    C
    `(a-b)g=(a_(1)-b_(1))g_(1)`
    D
    none of these

  • The pair of lines joining origin to the points of intersection of the two curves ax^2+2hxy+by^2+2gx=0 and a'x^2+2h'xy+b'y^2+2g'x=0 will be at right angles, if

    A
    `(a'+b')g'=(a+b)g`
    B
    `(a+b)g'=(a'+b')g
    C
    `h^2-ab = h'^2 - a'b'`
    D
    `a+b+h^2=a'b'+h^2`

  • The pair of lines joining origin to the points of intersection of the two curves ax^2+2hxy+by^2+2gx=0 and a'x^2+2h'xy+b'y^2+2g'x=0 will be at right angles, if

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    `(a'+b')g'=(a+b)g`
    B
    `(a+b)g'=(a'+b')g
    C
    `h^2-ab = h'^2 - a'b'`
    D
    `a+b+h^2=a'b'+h^2`

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