Samyak Jain
Last Activity: 6 Years ago
Compare x2/4 + y2/9 = 1 with x2/c2 + y2/d2 = 1.
c = 2, d = 3, e2 = 1 – c2/d2 = 1 – 4/9 = 5/9 (e = eccentricity of ellipse)
S(0,de) i.e. S(0,3e) , S’(0,–de) i.e. S’(0,–3e)
Take a point P(2cos
,3sin
) on the ellipse. Equation of tangent at P is T=0.
x.2cos
/4 + y.3sin
/9 = 1
xcos
/2 + ysin
/3 = 1
3cos
x + 2sin
y – 6 = 0 is the equation of tangent.
Length of perpendicular from S(0,3e) on the line is a = |0 + 6esin
– 6|/
i.e. a = 6(1 – esin
)/
[
esin
(0,1)]
Length of perpendicular from S’(0,–3e) on the line is c = |0 – 6esin
– 6|/
i.e. c = 6(1 + esin
)/
ac = 36(1 – e
2 sin
2)/(
) = 36[1 – (5/9)sin
2] / [9(1 – sin
2) + 4sin
2)]
= 36[(9 – 5sin
2)/9] / [9 – 9sin
2 + 4sin
2)]
ac = 4(9 – 5sin2) / (9 – 5sin2) = 4 Now,
|2x| dx =
|2x| + |2(-x)| dx =
4x dx
= 2
![\small \[x^2]_{0}^{ac}](https://latex.codecogs.com/gif.latex?%5Cdpi%7B100%7D%20%5Csmall%20%5C%5Bx%5E2%5D_%7B0%7D%5E%7Bac%7D)
= 2(ac)
2 = 2.(4)
2 = 32
