Eliminating $a,b$ from the system $2b+3a=ab$, $x_1=\frac{ab^2}{a^2+b^2}$ and $y_1=\frac{a^2b}{a^2+b^2}$

Question

I am trying to find the locus of foot of perpendicular drawn from the origin to a variable line passing through a fixed point $(2,3)$.

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My Attempt

Let the line be $bx+ay-ab=0$ where $a$ and $b$ are the axes intercepts. From the information that $(x_1,y_1)$ is foot of perpendicular we have the following:

$$\boxed{x_1=\dfrac{ab^2}{a^2+b^2}} \ \boxed{y_1=\dfrac{a^2b}{a^2+b^2}}$$

Also since the line passes through $(2,3)$ we have $2b+3a=ab$.

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Now I am not able to eliminate $a,b$. Since the varying parameters are only two and we have three equations, I am not short on equations. I just don't seem to know the way to manipulate the expressions that gets me rid of $a,b$. Any hints are appreciated. Thanks

Answer 1

The intercept form of equation of a line might not have been the best choice for this problem. There might be some clever insight that allows the two parameters to be eliminated with finesse, but here’s a way to grind it out directly.

First, solve for, say, $b$ in terms of $a$: $$b = {3a\over a-2}, a\ne2,$$ substitute and simplify to get (after some tedious calculation) $$x = {9a\over a^2-4a+13}, y = {3a^2-6a\over a^2-4a+13}.$$ To eliminate $a$, we need for the polynomials $(a^2-4a+13)x-9a$ and $(a^2-4a+13)y-(3a^2-6a)$ to vanish simultaneously, so we find their common roots by computing their resultant $$\begin{vmatrix}a^2x & -(4x+9)a & 13x & 0 \\ 0 & a^2x & -(4x+9)a & 13x \\ (y-3)a^2 & (6-4y)a & 13y & 0 \\ 0 & (y-3)a^2 & (6-4y)a & 13y \end{vmatrix} = 1053a^4(x^2-2x+y^2-3y).$$ This must vanish for all values of $a$, giving us the equation $$x^2-2x+y^2-3y=0$$ for the locus. This is a circle centered at $(1,3/2)$, which not coincidentally is the midpoint of the origin and $(2,3)$.

However, this equation of this locus can be written down more or less directly, without computing the foot of the perpendicular, or even an equation of a line through $(2,3)$ explicitly. We have the condition that the line segments from the point $(x,y)$ to the origin and to the fixed point $(2,3)$ must be orthogonal. Expressing this condition as the dot product of two vectors yields $$(x,y)\cdot(x-2,y-3)=0,$$ which is obviously equivalent to the equation derived above.

Answer 2

A line passing through (2,3) can be written as $y-3=m(x-2)$. The perpendicular from the origin can be written as $my=-x$

Since the intersection of two lines is the locus, substituting $(h,k)$ from the first equation $\frac{k-3}{h-2}=m, m=-\frac{h}{k}$. Hence $k(k-3)+h(h-2)=0$. The locus of the point is therefore, $x^2-2x+y^2-3y=0$

Answer 3

Locus will be circle with line joining origin O(0,0) and given point (2,3) as one of its diameter.$$ $$ Method is ::: let foot of perpendicular from origin on variable line L through (2,3) be point P(h,k). $$\therefore slope\,\, m_l=(\frac{k-3}{h-2})$$ $$Also \,\, slope m_{OP}=(\frac{k}{h})$$ Since OP is perpendicular to line L Hence $$m_l.m_{OP}=-1$$ $$\therefore (\frac{k-3}{h-2})(\frac{k}{h})=-1$$ $$\therefore (k-3)(k)+(h-2)(h)=0$$ Replace k with y and h with X Hence locus is $$x^2+y^2-2x-3y=0$$