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normal at 4 points (x1,y1),(x2,y2),(x3,y3),(x4,y4) of an ellipse are concurrent then prove (x1+x2+x3+x4)(1/x1+1/x2+1/x3+1/x4)=4

normal at 4 points (x1,y1),(x2,y2),(x3,y3),(x4,y4) of an ellipse are concurrent then  prove (x1+x2+x3+x4)(1/x1+1/x2+1/x3+1/x4)=4

Grade:11

1 Answers

mycroft holmes
272 Points
7 years ago
If the foot of the normal on the ellipse is (a \cos \theta, b \sin \theta), then the equation of the normal is ax \sec \theta - by \csc \theta = a^2-b^2
 
Say it passes through the point (h, k), then we have ah \sec \theta - bk \csc \theta = a^2-b^2

Let A = ah, B = a^2-b^2, C= bk,
 
Then we have (A \sec \theta - B) = C \csc \theta, squaring and writing RHS purely in terms of secant, we get
 
A^2 \sec^2 \theta -2AB \sec \theta+ B^2 = C^2 \frac{\sec^2 \theta}{\sec^2 \theta -1} 
 
which after clearing of denominators, we get a quartic 
 
A^2 \sec^4 \theta -2AB \sec^3 \theta+ (B^2-A^2-C^2) \sec^2 \theta + 2AB \sec \theta - B^2 = 0
Hence \sum \sec \theta_i = \frac{2AB}{B^2} = \frac{2B}{A}, \sum \cos \theta_i = \sum \frac{1}{\sec \theta_i} = \frac{2A}{B}
 
Thus we obtain \sum \sec \theta_i \times \sum \cos \theta_i = \left( \frac{2A}{B}\right ) \times \left( \frac{2B}{A}\right ) = \boxed{4}

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