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Given an ellipsoid with the formula: $$\mathcal{I}: \ \ \frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}{9}=1$$ Find the equation of the chord that passes through $M(2,1,-1)$ and is divided equally with this point (i.e $M$ lies on the center of the segment of the line inside the ellipsoid)

As we know one point on the line we have $$\frac{x-2}{l}=\frac{y-1}{m}=\frac{z+1}{n}$$ where $(l,m,n)$ is the direction vector. As $M$ is right on the center we have two points $A(a_1,a_2,a_3)$ and $B(b_1,b_2,b_3)$ such that $\frac{A+B}{2}=M$ and $A,B \in \mathcal{I}$. Then $(l,m,n)=(b_1-a_1,b_2-a_2,b_3-a_3)$.

From $A,B \in \mathcal{I}$, we may obtain that $$\frac{4}{25}l+8m-\frac{2}{9}n=0$$ However still I can't complete the task, I can't see where the right track is... Any help is aprreciated.

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  • $\begingroup$ You have 3 equations and 3 unknown. Solve the system. $\endgroup$
    – Moti
    Commented May 31, 2020 at 19:22
  • $\begingroup$ The intersections are calculated and than the distance. $\endgroup$
    – Moti
    Commented May 31, 2020 at 19:26
  • $\begingroup$ The result seems to be a closed line on the ellipsoid. $\endgroup$
    – Moti
    Commented May 31, 2020 at 19:34
  • $\begingroup$ @Moti you mean system of $A\in \mathcal{I}, B \in \mathcal{I}, AM=MB$? $\endgroup$
    – VIVID
    Commented May 31, 2020 at 19:35
  • $\begingroup$ Yes, as starter. You get x, y, and z in terms of l, m, n $\endgroup$
    – Moti
    Commented May 31, 2020 at 19:36

1 Answer 1

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Aside from what is likely a typo in your last equation, you’re on the right track. That last equation should be $$\frac4{25}l+\frac18m-\frac29n=0.$$

So far, you’ve determined that the chord must lie on the plane $\frac4{25}x+\frac18y-\frac29z={1201\over1800}$. From here, we can see that there isn’t a unique solution to the problem. The intersection of this plane turns out to be an ellipse centered at $M$, so every chord through $M$ that lies on this plane satisfies the conditions of the problem.

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  • $\begingroup$ Should the constant term be $\frac{1701}{1800}$ in the equation of the plane? If yes, then I got you. Thanks. $\endgroup$
    – VIVID
    Commented Jun 1, 2020 at 7:39
  • $\begingroup$ If we conclude all, then the intersection of the ellipsoid and this plane is a circle with the center at the point $M$? $\endgroup$
    – VIVID
    Commented Jun 1, 2020 at 7:40
  • $\begingroup$ @VIVID The right-hand side of the plane’s equation is $(4/25,1/8,-2/9)\cdot(2,1,-1)=1201/1800$. You can conclude that $M$ lies at the center of some conic on the plane, and moreover that it’s an ellipse, but it’s not a circle. Just compute the distances to the ellipsoid in a few different directions along the plane to see that they’re not all equal. $\endgroup$
    – amd
    Commented Jun 1, 2020 at 18:36

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