Let M(x, y), then: N(x, 0)
AN=|x+a|,NB=|x-a|,MN? =y?
So: |x+a|*|x-a|=ky?
Namely: |x? -a? |=ky?
This is the trajectory equation of point m.
When (1)|x|≦a, the equation is: a? -x? =ky?
Namely: x? +ky? =a? (k & gt0)
①0 & lt; K< is in 1, and the equation represents an ellipse with AB as its short axis;
② When k = 1, the equation represents a circle with a diameter of AB;
③k & gt; 1, where the equation represents an ellipse with AB as its long axis;
(2)|x| > When a, the equation is: x? -a? =ky? (k & gt0)
Namely: x? -ky? =a?
Therefore, at this time, the equation represents a hyperbola with AB as the real axis;
All in all,
When 0
When k= 1, the trajectory of m is: a circle with AB as its diameter and a hyperbola with AB as its real axis;
When k> is at 1, the trajectory of m is: an ellipse with AB as its long axis and a hyperbola with AB as its real axis;
Have fun! I hope I can help you. If you don't understand, please ask. I wish you progress in your study! O(∩_∩)O