Find the locus... (read on)?

Question

AB is a variable line sliding between the axes in such a way that A is on the x-axis and B is on the y-axis. P is a variable point on AB such that PA = a, PB = b and AB = a + b.

Find the locus of P.

Answer 1

Let P be (h,k) A be (d,0) and B be (0, e).

Therefore h = bd/(a+b) and k = ae/(a+b)

=> d = h(a+b)/b and e = k(a+b)/a

   d^2 = h^2(a+b)^/b^2 and e^2 = k^2(a+b)^2/a^2

     d^2 + e^2 = h^2(a+b)^2/b^2 + k^2(a+b)^2/a^2

    => (a+b)^2 = h^2(a+b)^2/b^2 + k^2(a+b)^2/a^2

       (By pythagorus theorm d^2 + e^2 = (a+b)^2)

    => 1 = h^2/b^2 + k^2/a^2

Changing h to x and k to y we get the locus of P as

          x^2/b^2 + y^2/a^2 = 1

which is the equation of an ellipse.

Answer 2

It will be an ellipse with center at (0,0), with major and minor axes 2a and 2b (not necessarily respectively; it depends which one is bigger)

the equation will be: x^2 / b^2 + y^2 / a^2 = 1

You can sketch a few cases to convince yourself of this...

edit: when A is at (-AB , 0), then B will be at the origin and P will be at (-b , 0)

when A is at (+AB,0), then B will still be at the origin, and P will be at (b,0)

When B is at (0,AB), then A will be at the origin, and P will be at (0,a)

when B is at (0,-AB), then A will still be at the origin, and P will be at (0,-a)

(that's the part where sketching it comes in handy!...)

The locus will be centered at the origin (0,0), giving the numerators of the equation.

Since the axis in the x-direction goes from (-b,0) to (b,0), that means that the semi-axis in the x-direction is b, which means that b^2 goes under x^2

Similarly, the axis in the y-direction goes from (0,a) to (0,-a), a distance of 2a.  The semi-axis in the y-direction is a, which means that a^2 goes under y^2