# equation of a locus

• Nov 28th 2009, 09:14 PM
ipokeyou
equation of a locus
Can you help me solve this please?
Thank you.

A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axes respectively, find the locus of the mindpoint of the ladder.
• Nov 28th 2009, 09:50 PM
mr fantastic
Quote:

Originally Posted by ipokeyou
Can you help me solve this please?
Thank you.

A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axes respectively, find the locus of the mindpoint of the ladder.

Is the ladder meant to be sliding?
• Nov 28th 2009, 11:06 PM
earboth
Quote:

Originally Posted by ipokeyou
Can you help me solve this please?
Thank you.

A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axes respectively, find the locus of the mindpoint of the ladder.

1. Draw the ladder in several different positions and determine the midpoint of the ladder.

2. Connect the midpoints by a smooth curve.

3. Let F denote the footpoint of the ladder with F(t, 0) and T the top of the ladder with $\displaystyle T(0, \sqrt{36-t^2})$. Then the midpoint of the ladder is
$\displaystyle M\left(\frac12 t\ ,\ \frac12 \sqrt{36-t^2} \right)$

Keep in mind that $\displaystyle 0 \leq t \leq 6$

4. That means the curve of all midpoints is described by the parametric equation:
$\displaystyle \left|\begin{array}{l}x=\frac 12 t\ ,\ 0 \leq t \leq 6 \\ y = \frac12 \sqrt{36-t^2}\end{array} \right.$

5. Solve the first equation for t and plug in this term into the 2nd equation. Re-arrange the equation until you have the standard equation of a circle around the origin with radius 3.