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

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

Grade:12

1 Answers

Samyak Jain
333 Points
5 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|/\sqrt{(3cos\theta)^2 + (2sin\theta)^2}
i.e. a = 6(1 – esin\theta)/\sqrt{9cos^2\theta + 4sin^2\theta}       [\because esin\theta \epsilon (0,1)]
Length of perpendicular from S’(0,–3e) on the line is c = |0 –  6esin\theta – 6|/\sqrt{(3cos\theta)^2 + (2sin\theta)^2}
i.e. c = 6(1 + esin\theta)/\sqrt{9cos^2\theta + 4sin^2\theta}
 
ac = 36(1 – e2 sin2\theta)/(\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, \int_{-ac}^{ac} |2x| dx = \int_{0}^{ac}|2x| + |2(-x)| dx = \int_{0}^{ac}4x dx
                            = 2\small \[x^2]_{0}^{ac}  =  2(ac)2  =  2.(4)2
                            = 32

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