The length of perpendicular from the foci S and S’ on any tangent to the ellipse x2/4+y2/9=1 are a and c respectively then find value of ∫{2x}dx limits -ac to ac

Samyak Jain
333 Points
4 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$\theta$,3sin$\theta$) on the ellipse. Equation of tangent at P is T=0.
x.2cos$\theta$/4 + y.3sin$\theta$/9 = 1  $\Rightarrow$ xcos$\theta$/2 + ysin$\theta$/3 = 1
3cos$\theta$ x + 2sin$\theta$ y – 6 = 0 is the equation of tangent.
Length of perpendicular from S(0,3e) on the line is a =  |0 + 6esin$\theta$ – 6|/$\dpi{80} \sqrt{(3cos\theta)^2 + (2sin\theta)^2}$
i.e. a = 6(1 – esin$\theta$)/$\dpi{80} \sqrt{9cos^2\theta + 4sin^2\theta}$       [$\dpi{80} \because$ esin$\dpi{100} \theta$ $\dpi{100} \epsilon$ (0,1)]
Length of perpendicular from S’(0,–3e) on the line is c = |0 –  6esin$\theta$ – 6|/$\dpi{80} \sqrt{(3cos\theta)^2 + (2sin\theta)^2}$
i.e. c = 6(1 + esin$\theta$)/$\dpi{80} \sqrt{9cos^2\theta + 4sin^2\theta}$

ac = 36(1 – e2 sin2$\theta$)/($\dpi{80} \9cos^2\theta + 4sin^2\theta$) = 36[1 – (5/9)sin2$\theta$] / [9(1 – sin2$\theta$) + 4sin2$\theta$)]
= 36[(9 – 5sin2$\theta$)/9] / [9 – 9sin2$\theta$ + 4sin2$\theta$)]
ac = 4(9 – 5sin2$\theta$) / (9 – 5sin2$\theta$)  =  4
Now, $\dpi{80} \int_{-ac}^{ac}$ |2x| dx = $\dpi{80} \int_{0}^{ac}$|2x| + |2(-x)| dx = $\dpi{80} \int_{0}^{ac}$4x dx
= 2$\dpi{100} \small \[x^2]_{0}^{ac}$  =  2(ac)2  =  2.(4)2
= 32