Let S be the focus, ZK the directrix and e the eccentricity of the ellipse whose equation is required.
Draw SK perpendicular from S on the directrix. Divide SK internally and externally at A and A’ (on SK produced) respectively in the ratio e : 1.
Draw SK perpendicular from S on the directrix. Divide SK internally and externally at A and A’ (on SK produced) respectively in the ratio e : 1.
∴ ⇒ SA = e. AK … (1)
And
⇒
SA’ = e A’K … (2)
Since A and A’ are such points that their distances from the focus bear constant ratio e(< 1) to their respective distances from the directrix. Therefore these points lie on the ellipse.
Let AA’ = 2a and C be the mid - point of AA’.
Then CA = CA’ = a
Adding (i) and (ii), we get
SA + SA’ = e (AK + A’K)
⇒
2a = e (CK – CA + A’C + CK)
⇒
2a = 2e CK [∵ CA = A’C = a]
⇒
CK = a/e … (3)
Subtracting (i) from (ii), we get
SA’ - SA = e (A’K - AK)
⇒
(SC + CA’) – (CA - CS) = e (AA’)
⇒
2CS = 2ae
⇒
CS = ae
Now let us choose C as the origin. CAX as x-axis and a line are (ae, 0) and equation of the directrix ZK is z = a/e.
Let P(x, y) be any point on the ellipse. Join SP and draw PMZK. Then, by definition, we have
SP = e PM
⇒
SP² = e² PM²
⇒
x² (1 - e²) + y² = a² (1 - e²)
⇒
, Where b² = a² (1 - e²)
This is the standard equation of the ellipse.
Note:
e < 1 ⇒
1 - e² < 1 ⇒ a² (1 - e²) < a² ⇒
b² = a².
Tracing of the ellipse , when a > b:
We have , where a > b … (1)
∴ … (2)
And … (3)
We observe the following:
a. Symmetry: For every value
of x there are equal and opposite values of y. Similarly for every value of y there are equal and opposite values of x.
Thus, the curve is symmetric about both the axes.
b. Origin: The curve does
not pass through the origin.
c. Intersection with axes: The curve meets x axis at y = 0
Putting y = 0 (iii), we get x = ± a.
So the curve meets y-axis at A (a, 0) and A’ (-a, 0).
Putting x = 0 in (ii), we get y = ± b.
So the curve meets y-axis at B (0, b) and B’ (0, -b).
d. Region: if x > a or, x < -a, from (ii) we get imaginary values of y, therefore, there is no part of the curve to the right of A or to the left of A’.
If y > b or y < -b, from (iii) we get imaginary values of x. therefore, there is no part of the curve above B (0, b) or below B’ (0, -b).
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