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Standard Equations of an Ellipse

To obtain an equation of the ellipse in a simplified form, we place the centre of the ellipse at the origin and position the ellipse so that its major axis coincides with one of the coordinate axis, say x-axis. If c is the distance from one focus to the centre, then one focus will be at F1 = (-c, 0) and the other one at F2 = (c, 0). We take the constant distance (sum of distances of point P from the foci F1 and F2) to be 2a. We shall take a>c. If P(x, y) is any point on the ellipse (Figure given above.) then by definition of the ellipse
PF1 + PF2 = 2a

Squaring both the sides, we get
(x + c)2 + y2 = 4a2 - 4a

Squaring both the sides, we get

Dividing both the sides by a2(a2 - c2), we get
If we put b2 = a2 - c2 where b > 0 then the above equation reads as

Alternative Method
To obtain an equation of the ellipse in a simplified form from the alternative definition. We proceed as follows Let F1 be the focus and D1 be the directrix of the ellipse (Figure given below.). Let M1 be the foot of perpendicular from F1 to D1.

Divide F1M1 internally and externally at V1 and V2, respectively in the ratio e:1, so that



Thus by definition V1 and V2 lie on the ellipse.

Let O be the mid-point of V1V2 suppose V1V2 = 2a.
Then OV1 = OV2 = a.
From (1), we get F1 V1 = e V1M1 and F1V2 = e V2M1   ------------------(2)
Adding, we get F1V1 + F1 V2 = e(V1M1 + V2M1)
V1V2 = e[(OM1 - OV1) + (OM1 + OV2)]
2a = 2e(OM1)                           [OV1 = OV2 = a]
OM1 = .

Subtracting the relations in (2), we get F1V2 F1V1 = e V2M1 - e V1M1
(OF1 + OV2) (OV1 OF1) = e(V2M1 V1M1)
2OF1 = e (V1V2)                        [OV2 = OV1 = a]
OF1 = ae

Take O as the origin, OV1 as the x-axis and OY perpendicular to OV1 as y-axis.
Since F1 is to the left of the y-axis and OF1 = ae, the coordinates of F1 are ( ae, 0) and since OM1 = , the equation of the directrix D1 is .

Let P(x, y) be any point on the ellipse. Let M be the foot of perpendicular from P to D1. By definition of the ellipse
⇒ F1P2 = e2PM2 -------------------(3)

We have F1P2 = (x + ae)2 + y2 and PM = .

Substituting these values in (3), we get
x2 + 2xae + a2 e2 + y2 = e2x2 + 2aex + a2
x2 (1 e2) + y2 = a2(1 e2)
Dividing throughout by a2 (1 e2), we get
Since 0 < e < 1, a2 (1 e2) > 0. We take a2 (1 e2) = b2 (b > 0), so that the above equation becomes
   ------------------- (4)

Conversely, every equation of the form (4) represents an ellipse.
If P (x, y) satisfies the equation (4), then by retracing our steps we can show that PF1 = e PM, so that the point P (x, y) lies on the ellipse. Hence, the equation
represents an ellipse.

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