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A variable line $L$ passing through the point $B(2, 5)$ intersects the crossed lines $$2x^2 − 5xy + 2y^2 = 0$$ at $P$ and $Q$. Find the locus of the point $R$ on $L$ such that distances $BP$, $BR$, and $BQ$ are in harmonic progression.

The two crossed lines are $x=2y$ and $2x=y$. The line passing through $(2,5)$ is $\dfrac{y-5}{x-2}=m$.

On intersecting the variable line with $x=2y$ and $2x=y$, we get the point $P$ and $Q$ in form of $m$.

But I am confused. As $(h,k)$ represent the point $R$, $$\frac{k-5}{h-2}=m$$ We need to eliminate $m$ in order to find the locus, but not able to eliminate $m$.

Blue
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Samar Imam Zaidi
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  • "the line $2x^2−5xy+2y^2=0$", that is a parabola not a line! Why are you talking about the two lines $x=2y$ and $2x=y$? – Henrik supports the community Feb 01 '19 at 06:52
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    @Henrik: $2x^2-5xy+2y^2=0$ factors as $(x - 2 y) (2 x - y)=0$. It's a degenerate crossed-lines conic. (It's not a parabola, since $(-5)^2-4(2)(2)=9\neq 0$.) – Blue Feb 01 '19 at 06:58

2 Answers2

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Equation of line passing through point (2,5) is:

$$y=5+k(x-2)\tag{1}$$

We also know that this line interesects curve:

$$2x^2-5xy+2y^2=0\tag{2}$$

...in points $P(x_P, y_P)$ and $Q(x_Q, y_Q)$. To obtain coordinates of $P,Q$ just replace (1) into (2) to eliminate $y$. You get the following quadratic equation:

$$(50-40k+8k^2)+(-25+30k-8k^2)x+(2-5k+2k^2)x^2=0$$

with the following solutions:

$$x_P=\frac{2k-5}{k-2}, \quad x_Q=\frac{2(2k-5)}{2k-1}\tag{3}$$

From (1) and (3) you get:

$$y_P=\frac{2(2k-5)}{k-2}$$

$$y_Q=\frac{2k-5}{2k-1}$$

Segments $BP,BR,BQ$ must be in harmonic progression. This is equivalent to:

$$x_R-x_B=\frac{2}{\frac1{x_P-x_B}+\frac1{x_Q-x_B}}$$

$$y_R-y_B=\frac{2}{\frac1{y_P-y_B}+\frac1{y_Q-y_B}}$$

or:

$$x_R=\frac{20k-50}{10k-17}$$

$$y_R=\frac{34k-85}{10k-17}$$

Eliminate $k$ from the last two equations and you get:

$$y_R=\frac{17}{10}x_R$$

Saša
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Well, I solve this problem, but I have not much confidence that it is true.

However, I think the approach is almost right. Please check it, and if there is some false part, let me know.

First, I translate all objects $-2$ in the direction of $x$-axis, and $-5$ in the direction of $y$-axis.

Then $B$ become $B'(0,0)$, the origin is translated to $(-2,-5)$, and two line become $y=2x-1$, $y=\frac{1}{2}x-4$.

Assume $B'(0,0)$, $P'(t,2t-1)$, and $Q'(s,\frac{1}{2}s-4)$ lie on the one straight line.

Then the equation $t:2t-1 = s: \frac{1}{2}s-4$ holds.

Thus, $t(\frac{1}{2}s-4)=(2t-1)s$.

From this equation, we can get $s=\frac{-8}{3t-2}t$.

Thus We can say that $B'=(0,0)$, $P'=1\cdot(t,2t-1)$, and $Q'=\frac{-8}{3t-2} \cdot(t,2t-1)$.

Because the harmonic mean of $1$ and $\frac{-8}{3t-2}$ is $\frac{-16}{3t-10}$,

We can get $R'=\frac{-16}{3t-10} \cdot (t,2t-1) = \big(\frac{-16t}{3t-10},\frac{-16(2t-1)}{3t-10}\big)$

If we let $R'=(x,y)$, then $x=\frac{-16t}{3t-10}$, and $t=\frac{10x}{3x+16}$.

Then $y=\frac{-16(2t-1)}{3t-10}=\frac{17x-16}{10}$.

By translating this line $+2$ in the direction of $x$-axis, and $+5$ in the direction of $y$-axis,

$y-5=\frac{17(x-2)-16}{10}$, i.e., $y=\frac{17}{10}x$.

It is the desired locus.

with-forest
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