A line segment of length 12 moves with its ends always touching the coordinate axes. Find the equation on the graph of the point on the segment that is 4 units from the end in contact with the x - axis.
The title "Another one on ellipse" is the clue. Without playing yet, the locus of that particular point is an ellipse.
After some playing with the line segment and the two coordinate axes, yes, the locus is an ellipse.
If the line segment is vertical, the point is 4 units from the x-axis.
Sliding the line segment to the left, when it is horizontal, the point is 8 units from the y-axis.
Then, sliding the line segment down, when it is vertical again, the point is 4 units from the x-axis.
Then, sliding the line segment to the right, when it is horizontal again, the point is 8 units from the y-axis.
Then, sliding the line segment up, when it is vertical, it is in the same position before, the point is 4 units from the x-axis.
So, the ellipse is "horizontal" or the major axis is along the x-axis.
major axis = 8+8 = 16 units long.
minor axis = 4+4 = 8 units high.
A standard equation of an ellipse centered at the origin (0,0), whose major axis, 2a, is along the x-axis, and whose minor axis, 2b, is then along the y-axis is
(x^2)/(a^2) +(y^2)/(b^2) = 1
Therefore the equation of the locus of that particular point is:
(x^2)/(8^2) +(y^2)/(4^2) = 1
(x^2)/64 +(y^2)/16 = 1 -----------------answer.
Hello, ^_^Engineer_Adam^_^!
It took me several tries, but I found a painless approach . . .
A line segment of length 12 moves with its ends always touching the coordinate axes.
Find the equation on the graph of the point on the segment that is 4 units
from the end in contact with the x - axis.Code:| (0,b)* | * | * | * P(x,y) + - - - - - - - o | : * ------+---------------+-------*-- | (a,0)
Let the intercepts of the segment be $\displaystyle (a,0)$ and $\displaystyle (0,b).$
. . Then: .$\displaystyle a^2 + b^2\:=\:12^2$ [1]
From similar triangles we have: .$\displaystyle \begin{Bmatrix}x = \frac{2}{3}a\quad\Rightarrow\quad a = \frac{3}{2}x \\ \\
y = \frac{1}{3}b\quad\Rightarrow\quad b = 3y \end{Bmatrix}$
Substitute into [1]: .$\displaystyle \left(\frac{3}{2}x\right)^2 + (3y)^2\:=\:144\quad\Rightarrow\quad \frac{9x^2}{4} + 9y^2\;=\;144$
Divide by $\displaystyle 144\!:\;\;\boxed{\frac{x^2}{64} \,+ \,\frac{y^2}{16}\;=\;1}$ . . . ta-DAA!