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Math Help - Locus of foot of perpendicular from origin

  1. #1
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    Smile Locus of foot of perpendicular from origin

    Form the equation of straight lines joining the origin to the points of intersection of ax^2+2hxy+by^2+2gx+2fy+c=0 and px+qy+r=0 and write down the condition that these lines should be right angles.
    If this condition is satisfied, show that the equation of the locus of the foot of the perpendicular from the origin to the line px+qy+r=0 is
    (a+b)(x^2+y^2)+2gx+2fy+c=0

    I've done the first part. The answers I found are:
    x^2(ar^2-2gpr+p^2c)+2xy(hr^2-fpr-gqr+pqc)+y^2(br^2-2fqr+q^2c)=0
    and
    r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0
    How do I proceed with the next part?
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Form the equation of straight lines joining the origin to the points of intersection of ax^2+2hxy+by^2+2gx+2fy+c=0 and px+qy+r=0 and write down the condition that these lines should be right angles.
    If this condition is satisfied, show that the equation of the locus of the foot of the perpendicular from the origin to the line px+qy+r=0 is
    (a+b)(x^2+y^2)+2gx+2fy+c=0

    I've done the first part. The answers I found are:
    x^2(ar^2-2gpr+p^2c)+2xy(hr^2-fpr-gqr+pqc)+y^2(br^2-2fqr+q^2c)=0
    and
    r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0
    How do I proceed with the next part?
    It looks as though you are correct so far, and quite close to completing the question. The foot of the perpendicular from the origin to px+qy+r=0 is the point where that line meets the line qx-py=0. You can easily check that this is the point (x,y), where x = \frac{-pr}{p^2+q^2} and y = \frac{-qr}{p^2+q^2}. Notice that x^2+y^2 = \frac{r^2}{p^2+q^2}.

    Now divide the equation r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0 by p^2+q^2, and it should be obvious where to go from there.
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  3. #3
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    ok now i have a problem with the second term, how do i get rid of the p and q? Thanks!
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  4. #4
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    Quote Originally Posted by arze View Post
    ok now i have a problem with the second term, how do i get rid of the p and q? Thanks!


    Now divide the equation by

    \frac{r^2(a+b)}{(p^2+q^2)} - \frac{2r(gp + fq)}{(p^2+q^2)} + c = 0

    Now substitute the values of
    .

    and .

    And see what you get.
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