Locus of point M such that its projections onto two fixed lines are always at the same distance.

Question

A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is

$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$

What I have done: Suppose $P$ is any point on the locus, $A$ and $B$ be the feet of the perpendiculars, and $O$ is the origin. Then I want to show that

$$OP=\frac{AB}{\sin AOB}=\frac{c}{\sin \theta}$$

where $\theta$ is the angle between the two lines.

Note that the arc $PAO$ is a semi-circle, since $\angle PAO=90$. Similarly the arc $PBO$ is a semi-circle since $\angle PBO=90$. Also $\angle APB=\angle AOB$, as it is the angle of the same arc.

Now, $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. From here I can find $\sin \theta$ and $OP=\sqrt{x^2+y^2}$. Putting all these in,

$$OP=\frac{c}{\sin \theta}$$

will give me the equation of the locus. The problem is how to show that

$$OP=\frac{AB}{\sin AOB}.$$

Answer 1

Please note that $OAPB$ is cyclic quadrilateral and $OP$ is the diameter of the circle. Also both lines pass through origin.

So if $(x, y)$ is the coordinates of $P$, radius of the circle is $ \displaystyle r = \frac{\sqrt{x^2 + y^2}}{2}$.

If angle between two lines is $\theta$, the angle subtended by chord $AB$ on the center is $2\theta$. We then have,

$ \displaystyle AB = 2 r \sin \theta \implies \sin\theta = \frac{c}{ \sqrt{x^2+y^2}}$ (as $AB = c$)

Now if the equations of lines are $l_1: a_1 x + b_1 y = 0$ and $l_2: a_2 x + b_2 y = 0$

Multiplying both equations and equating to $ax^2 + 2hxy+b y^2 = 0$,

$a_1a_2 = a, \ b_1b_2 = b, \ a_1b_2 + a_2b_1 = 2h \tag1$

Also $ \displaystyle m_1 = - \frac{a_1}{b_1}, m_2 = - \frac{a_2}{b_2}$

So, $ \displaystyle \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{a_1 b_2 - a_2b_1}{a_1a_2 + b_1b_2}\right| = \left|\frac{ \sqrt{(a_1 b_2 + a_2b_1)^2 - 4 a_1 a_2 b_1 b_2}}{a_1a_2 + b_1b_2}\right|$

With help of equation $(1)$, $ \displaystyle \tan \theta = \left|\frac{ \sqrt{4h^2 - 4 ab}}{a + b}\right|$

Using $ \displaystyle \sin^2 \theta = \frac{\tan^2\theta}{1+\tan^2\theta} \ $, $ \ \displaystyle \frac{c^2}{x^2+y^2} = \frac{4h^2 - 4ab}{4h^2 + (a-b)^2}$

EDIT: seeing Jean Marie's nice diagrams, I realize adding an example to go with my answer may help visualize the situation better. I do not have such nice tools with me so my diagram is basic Desmos :)

Before we get to the example, using the formula we derived, it is easy to see that the locus is a circle centered at the origin. Now why origin? Simply because the question uses pair of lines intersecting at the origin. In other words, if the lines intersected at a different point, that point will be the center of the circle.

Let's take pair of lines given by $x^2 - 5xy + 4y^2 = 0$. So we have $a = 1, b = 4, h = - \frac{5}{2}$.

The lines are $x - y = 0, x - 4y = 0$.

So using the formula we derived earlier, the locus of the point in this example is,

$ \displaystyle x^2 + y^2 = \frac{34 c^2}{9}$. See the Desmos diagram for $c = 1$. From any point on this circle, if we draw perpendicular to the given pair of lines, the distance between the feet of both perpendicular will be $1$.

Answer 2

Here's a non-trigonometric take on this problem.

The conic $ax^2+2hxy+by^2=0$ gives the two solutions $y=m_\pm x$ where $bm_\pm=-h\pm\sqrt{h^2-ab}$. Given $P(x_0,y_0)$, the perpendicular foot $A(x,m_+x)$ minimises $AP^2=(x-x_0)^2+(m_+x-y_0)^2$. This is minimised at $(m_+^2+1)x^*=x_0+m_+y_0$ so $$A\left(\frac{x_0+m_+y_0}{m_+^2+1},\frac{m_+(x_0+m_+y_0)}{m_+^2+1}\right)\implies B\left(\frac{x_0+m_-y_0}{m_-^2+1},\frac{m_-(x_0+m_-y_0)}{m_-^2+1}\right).$$ without loss of generality. As $AB=c$ we have $c^2=(\alpha^2+\beta^2)(x_0^2+y_0^2)$ where $\alpha=\dfrac1{m_+^2+1}-\dfrac1{m_-^2+1}$ and $\beta=\dfrac{m_+}{m_+^2+1}-\dfrac{m_-}{m_-^2+1}$. Now \begin{align}\alpha^2+\beta^2&=\frac1{m_+^2+1}+\frac1{m_-^2+1}-\frac{2(m_+m_-+1)}{(m_+^2+1)(m_-^2+1)}\\&=\frac{(m_+-m_-)^2}{(m_+m_-)^2+m_+^2+m_-^2+1}\\&=\frac{4(h^2-ab)/b^2}{a^2/b^2+2(2h^2-ab)/b^2+1}\\&=\frac{4(h^2-ab)}{4h^2+(a-b)^2}\end{align} using Vieta so the locus is a circle centred at the origin of radius $\dfrac c2\sqrt{\dfrac{4h^2+(a-b)^2}{h^2-ab}}$.

Answer 3

As it is a geometry problem, I am reluctant to stay with computations without any drawing. This is what I did ; the drawings unveil unsuspected aspects of the issue. Here are two drawings that I will detail later on, in connection with a classical problem.

Fig. 1: The exceptional case where the two lines are othogonal. In this case, $r=c$ (see relationship (3)). The blue segments represent different positions of constant length line segment $M_1M_2$ (joining the feet of the perpendiculars, represented in red). We have here the classical representation of the sliding ladder problem.

Fig. 2: The general case with an angle $\alpha$ between the two lines ; here $\alpha = \pi/6$ giving a circle with radius $r=2c$ (see relationship (3)). This figure can be seen as an extension of the sliding ladder problem to the case of a non vertical wall.

First of all, the union of the two straight lines can be written WLOG under the form:

$$(x\sin \alpha - y\cos \alpha)(x\sin \beta - y\cos \beta)=0.\tag{1}$$

One can even assume (again WLOG, because the issue is invariant with respect to rotations) that angle $\beta$ is $0$ (i.e., it can be assumed that one of the straight lines is taken as abscissas axis) [see Fig. 2]. Relationship (1) is therefore equivalent to:

$$(x\sin \alpha - y\cos \alpha)y=0.\tag{2}$$

Comparison with expression $ax^2+2hxy+by^2=0$ gives

$$a \ = \ 0, \ \ \ \ h \ = \ \frac12 \sin \alpha, \ \ \ \ b \ = \ -\cos \alpha$$

With these new parameters, the equation of the circle to be found:

$$x^2+y^2 \ \ = \ \ \left(\frac{c}{2}\right)^2 \ \ \dfrac{4h^2+(a-b)^2}{h^2-ab}$$

becomes

$$x^2+y^2\overset{?}{=}r^2 \ \ \text{with} \ \ r:=\frac{c}{\sin \alpha}\tag{3}$$

Let us establish relationship (3).

Let $M(x,y)$ be the current point. Let $M_1$ and $M_2$ be its orthogonal projections on the horizontal and slanted axes resp.

• Evidently: $M_1(x,0).$

• The slant axis being directed by unit vector $\vec{u}=\binom{\cos \alpha}{\sin \alpha}$, we have: $M_2=(\underbrace{\vec{OM}.\vec{u}}_s) \vec{u}=\binom{s \cos \alpha}{s \sin \alpha},$ where $s:=(x \cos \alpha + y\sin \alpha)$ is the dot product.

Expressing now constraint $(M_1M_2)^2=c^2$:

$$(x-s \cos \alpha)^2+(s \sin \alpha)^2=c^2$$

which, expanded, gives relationship (3), as desired.

We have thus provided another coordinate geometry proof.

But, as announced, there is more to be said on the side of geometric understanding.

There exists a connection with a classical cinematics issue that many among us have been working on, the sliding ladder problem (fig. 1). The envelope of the ladder is an astroid and the circle is the locus of all the instantaneous centers of rotation of the ladder while it is sliding (see the very interesting article here).

For the general case, that can be called the sliding ladder problem with slanted wall, I have found hardly one reference.

The circle remains the locus of the center of intantaneous rotation, but the envelope is no longer an astroid. I don't know if it has been investigated. [This could become a new question...]

Answer 4

To add further graphic context to the invariant radius OP, I indicated two positions of the sliding ladder generating an " Oblique Astroid " to Math Love's sketch in following way:

$$ OP=\frac{c}{\sin \alpha} \quad= \text{ Radius of locus = Circum-diameter of cyclic quadrilateral}. $$

Intersections of the green circle through center ( the line pair intersection point O of the inclined pair of lines) enclose a constant chord length =$c.$

When $\alpha$ is acute the painter works in a narrow uncomfortable sector and with supplementary obtuse angle he is more comfortable.

The locus can be found independently as follows.

Let two lines AO, OB cut at O, including an angle $\alpha$ in between them.

By a converse of Euclid's proposition, if four points( A,O, B, C' ) are placed such that the exterior angle $ \alpha$ at O equals angle in the alternate segment, then they are concyclic. Shift C' to C with equal angle $\alpha$ in the same segment so that angle made at B is $90^{\circ}$. Diameter AC now contains center of the circle.

By Sine Rule $ d= \dfrac{c}{\sin \alpha}= AC $ the circum-diameter, that is also a constant because $ \alpha, c $ are both constant, leaving out O as a free fulcrum to rotate all rigid circles through it, always enclosing ladder length $ c= AB $ standing between the fixed first two lines points of intersection.

Answer 5

To easily prove that the equation of locus whose equation is recognizable to be a circle of radius R we are given

$$4(x^2+y^2)(h^2-ab)=4 R^2 (h^2-ab)=c^2(4h^2+(a-b)^2) $$

i.e., we are given

$$ \frac{R^2}{c^2}=\frac{(4h^2+(a-b)^2)}{4(h^2-ab)} \tag 1 $$

Consider the special case when the moving point is directly on one of the pair of straight lines configured in this simplest way:

$$ \frac {c}{R} = \sin \alpha \tag 2 $$

Angle included between straight line pair is known, given by:

$$ \tan \alpha =\frac{\sqrt{h^2-ab}}{(a+b)/2} \tag 3 $$

Elimination of $\alpha$ between equations (2) and (3) results in (1)... done !