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**

**Definition of Ellipse****Standard Equation of an Ellipse****Two Foci and Two Directrices of the Ellipse****Vertex of the Ellipse****Centre of the Ellipse****Major and Minor Axes of the Ellipse****Latus Rectum of the Ellipse****Position of a Point with respect to the Ellipse****Ellipse Formulae****Focal Distance of a Point on the Ellipse****Problems on Ellipse**

**11 and 12 Grade Math**__From Standard Equation of an Ellipse____ to HOME PAGE__

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.