The Three e e 's

Geometry Level 5

Consider an ellipse on the Cartesian Plane. Its x-intercepts are 2 \sqrt{2} and − 2 -\sqrt{2} , and its e = 1 2 e=\frac{1}{\sqrt{2}} . Now, it intersects with y = e x y=e^x at two points. Let these points be ( x 1 , y 1 ) , ( x 2 , y 2 ) (x_1,y_1),(x_2,y_2) , with x 1 < x 2 x_1<x_2 . Now, consider the ellipse to be solid, and let acceleration due to gravity act in the direction of the negative y-axis. Let a particle be projected from ( x 1 , y 1 ) (x_1,y_1) to ( x 2 , y 2 ) (x_2,y_2) . The particle collides with the ellipse with an e = 1 e=1 . The particle bounces off the ellipse, and lands back onto the ellipse at ( x 3 , y 3 ) (x_3,y_3) . Find x 3 x_3 to 4 4 decimal places.

In this question, the first e e refers to the eccentricity of the ellipse. The second e e refers to the Euler's Constant ≈ 2.7 … \approx2.7\ldots . The third e e refers to the coefficient of restitution of collision.


The answer is 1.3675.

Nanayaranaraknas Vahdam by Nanayaranaraknas Vahdam , 22, India

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1 solution

WE CAN EASILY GET AN INTERSECTION OF GIVEN ELLIPSE AND e x e^x , i.e. (0,1),

So the normal at that point to ellipse is y-axis ,

if the other point of intersection of ellipse and e x e^x be (x,y),

Then , x 3 x_{3} will be equal to − x -x because it is given that e=1 (coefficient of restitution),

So , substituting y y = e x e^x in ellipse equation which is x 2 x^2 + 2 y 2 y^2 = 2 ,

We get x 2 x^2 - 2 + e ( 2 x ) e^(2x) = 0 ,

Solving we get x x = -1.3675,

So x 3 x_{3} = -x = 1.3675 1.3675 .

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