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

• 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 

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