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
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.
May 18, 24 02:59 PM
May 12, 24 06:28 PM
May 12, 24 06:23 PM
May 12, 24 06:09 PM
May 12, 24 04:59 PM
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.