Current location - Training Enrollment Network - Mathematics courses - Senior high school mathematics simple trajectory equation solution master.
Senior high school mathematics simple trajectory equation solution master.
Let AB=2a, and establish a plane rectangular coordinate system with the midpoint of A and B as the origin, then: A(-a, 0), B(a, 0).

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