# Locus of foot of perpendicular from origin

#### arze

Form the equation of straight lines joining the origin to the points of intersection of $$\displaystyle ax^2+2hxy+by^2+2gx+2fy+c=0$$ and $$\displaystyle px+qy+r=0$$ 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 $$\displaystyle px+qy+r=0$$ is
$$\displaystyle (a+b)(x^2+y^2)+2gx+2fy+c=0$$

I've done the first part. The answers I found are:
$$\displaystyle x^2(ar^2-2gpr+p^2c)+2xy(hr^2-fpr-gqr+pqc)+y^2(br^2-2fqr+q^2c)=0$$
and
$$\displaystyle r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0$$
How do I proceed with the next part?
Thanks!

#### Opalg

MHF Hall of Honor
Form the equation of straight lines joining the origin to the points of intersection of $$\displaystyle ax^2+2hxy+by^2+2gx+2fy+c=0$$ and $$\displaystyle px+qy+r=0$$ 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 $$\displaystyle px+qy+r=0$$ is
$$\displaystyle (a+b)(x^2+y^2)+2gx+2fy+c=0$$

I've done the first part. The answers I found are:
$$\displaystyle x^2(ar^2-2gpr+p^2c)+2xy(hr^2-fpr-gqr+pqc)+y^2(br^2-2fqr+q^2c)=0$$
and
$$\displaystyle r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0$$
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 $$\displaystyle px+qy+r=0$$ is the point where that line meets the line $$\displaystyle qx-py=0$$. You can easily check that this is the point $$\displaystyle (x,y)$$, where $$\displaystyle x = \frac{-pr}{p^2+q^2}$$ and $$\displaystyle y = \frac{-qr}{p^2+q^2}$$. Notice that $$\displaystyle x^2+y^2 = \frac{r^2}{p^2+q^2}$$.

Now divide the equation $$\displaystyle r^2(a+b)-2r(gp+fq)+c(p^2+q^2)=0$$ by $$\displaystyle p^2+q^2$$, and it should be obvious where to go from there.

arze

#### arze

ok now i have a problem with the second term, how do i get rid of the p and q? Thanks!

#### sa-ri-ga-ma

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

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

Now substitute the values of
.

and .

And see what you get.

arze
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