# Standard Equation of an Ellipse

We will learn how to find the standard equation of an ellipse.

Let S be the focus, ZK the straight line (directrix) of the ellipse and e (0 < e < 1) be its eccentricity. From S draw SK perpendicular to the directrix KZ. Suppose the line segment SK is divided internally at A and externally at A' (on KS produced) respectively in the ratio e : 1.

Therefore, $$\frac{SA}{AK}$$ = e : 1

$$\frac{SA}{AK}$$ = $$\frac{e}{1}$$

⇒ SA = e ∙ AK ...................... (i) and

$$\frac{SA'}{A'K}$$ = e : 1

$$\frac{SA'}{A'K}$$ = $$\frac{e}{1}$$

⇒ SA' = e ∙ A'K ...................... (ii)

We can clearly see that the points A and A'' lies on the ellipse since, their distance from the focus (S) bear constant ratio e (< 1) to their respective distance from the directrix.

Let C be the mid-point of the line-segment AA'; draw CY perpendicular to AA'.

Now, let us choose C as the origin CA and CY are chosen as x and y-axes respectively.

Therefore, AA' = 2a

A'C = CA = a.

Now, adding (i) and (ii) we get,

SA + SA' = e (AK + A'K)

AA' = e (CK - CA + CK + CA')

2a = e (2CK - CA + CA')

2a = 2e CK,  (Since, CA = CA')

CK = $$\frac{a}{e}$$ ...................... (iii)

Similarly, subtracting (i) from (ii) we get,

SA' - SA = e (KA' - AK)

(CA' + CS) - (CA - CS) = e . (AA')

2CS = e 2a, [Since, CA' = CA]

CS = ae ...................... (iv)

Let P (x, y) be any point on the required ellipse. From P draw PM perpendicular to KZ and PN perpendicular to CX and join SP.

Then, CN = x, PN = y and

PM = NK = CK - CN = $$\frac{a}{e}$$ – x, [Since, CK = $$\frac{a}{e}$$] and

SN = CS - CN = ae - x, [Since, CS = ae]

Since the point P lies on the required ellipse, Therefore, by the definition we get,

$$\frac{SP}{PM}$$ = e

SP = e PM

SP$$^{2}$$ = e$$^{2}$$ . PM$$^{2}$$

or  (ae - x)$$^{2}$$ + (y - 0)$$^{2}$$ = e$$^{2}$$[$$\frac{a}{e}$$ - x]$$^{2}$$

⇒ x$$^{2}$$(1 – e$$^{2}$$) + y$$^{2}$$ = a$$^{2}$$(1 – e$$^{2}$$)

$$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{a^{2}(1 - e^{2})}$$ = 1

$$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{a^{2}(1 - e^{2})}$$ = 1

Since 0 < e < 1, hence a$$^{2}$$(1 - e$$^{2}$$) is always positive; therefore, if a$$^{2}$$(1 - e$$^{2}$$) = b$$^{2}$$, the above equation becomes,  $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1.

The relation $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 is satisfied by the co-ordinates of all points P (x, y) on the required ellipse and hence, represents the required equation of the ellipse.

The equation of an ellipse in the form $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 is called the standard equation of the ellipse.

Notes:

(i) b$$^{2}$$ < a$$^{2}$$, since e$$^{2}$$ < 1 and b$$^{2}$$ = a$$^{2}$$(1 - e$$^{2}$$)

(ii)  b$$^{2}$$ = a$$^{2}$$(1 – e$$^{2}$$)

$$\frac{b^{2}}{a^{2}}$$ = 1 – e$$^{2}$$, [Dividing both sides by a$$^{2}$$]

e$$^{2}$$ = 1 - $$\frac{b^{2}}{a^{2}}$$

e = $$\sqrt{ 1 - \frac{b^{2}}{a^{2}}}$$, [taking square root on both sides]

Form the above relation e = $$\sqrt{ 1 - \frac{b^{2}}{a^{2}}}$$, we can find the value of e when a and b are given.

● The Ellipse