need  equation of the ellipse with foci (7, 0), (-7, 0) and co-vertices (0, 3), (0, -3)?

The general equation of a centralised ellipse is x2/a2+y2/b2=1 where a and b are the lengths of the sem-axes.

Midway between the foci is the centre of the ellipse, which in this case must be (0,0) because it's the midpoint of (-7,0) and (7,0) and the midpoint of the covertices. Also, the foci lie on the x-axis so a is the distance between the centre of the ellipse (0,0) and the horizontal vertices on the major axis. The distance between these vertices is 2a. Let c be the focal length, that is the distance between each focus and the centre of the ellipse.

The covertices must lie on the ellipse so 9/b2=1, and b=3 (plug in x=0 and y=±3).

If P is a point on the ellipse then PF1 and PF2 are the distances between P and each focus. Their sum is always constant no matter where P is. At a co-vertex we have the apex of an isosceles triangle, so if P is a co-vertex then PF1=PF2. The extremes of the horizontal vertices. If O is the centre of the ellipse then PF22= PF12=PO2+F1O2 (Pythagoras). But PO=3 and F1O=F2O=7.

PF22= PF12=9+49=58. This implies that PF2= PF1=√58 and their sum is 2√58.

This sum is the same no matter where P is, so if P is at vertex (a,0), then PF1+PF2=2√58. But PF1=a-c and PF2=a+c (simple geometry), PF1+PF2=2a, making a=√58.

The equation of the ellipse is x2/58+y2/9=1 in standard form.

by Top Rated User (1.1m points)