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An ellipse has its center at the origin and its minor axis is along the x-axis. If the distance between its foci is equal to the length of its minor axis and the length of its latus rectum is 4, then which of the following points lies on the ellipse?$$(2,2\sqrt2)/(2,\sqrt2)/(\sqrt2,2\sqrt2)/(2\sqrt2,2)$$

My attempt: Let $a,b, e$ be the length of minor axis, length of major axis and eccentricity, respectively. So, $2be=2a\implies a=be$. Also, $\frac{2a^2}{b}=4\implies\frac{b^2e^2}{b}=2$. Don't know how to proceed next.

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Hint:

Use the fact that $$e=\sqrt{1-\frac{a^2}{b^2}} $$

Since $a=be$, $$e=\sqrt{1-e^2} \implies e=\frac{1}{\sqrt 2}$$

Now use this value to get $a$ and $b$, and find which point satisfies the equation of the ellipse.

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  • $\begingroup$ I guess, in this case, it would be $\frac{a^2}{b^2}$. I had used that, in vain. $\endgroup$
    – aarbee
    May 6 '20 at 11:44
  • $\begingroup$ What is the equation of the ellipse you are assuming? $\endgroup$
    – Tavish
    May 6 '20 at 11:45
  • $\begingroup$ $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $\endgroup$
    – aarbee
    May 6 '20 at 11:50
  • $\begingroup$ @Ramit You’ll find my edit useful. $\endgroup$
    – Tavish
    May 6 '20 at 12:12
  • $\begingroup$ Thanks, it's appearing so easy now! $\endgroup$
    – aarbee
    May 6 '20 at 12:14
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You really only need to find the semiminor axis length in order to solve this problem.

Letting $a$ be the semiminor axis length as you did, since this is also equal to half of the distance between the foci, then the sum of distances from the foci to any point on the ellipse is equal to $2\sqrt2a$. The semi-latus rectum is $2$, so the distance from an endpoint of the latus rectum to the other focus is, via the Pythagorean theorem, equal to $\sqrt{2^2+(2a)^2}=2\sqrt{1+a^2}$. Combining these two facts, we have $2+2\sqrt{1+a^2}=2\sqrt2a$, therefore $a=2\sqrt2$.

Now, examining the four given points, it’s easy to see that the first one is an endpoint of the latus rectum in the first quadrant, so the second and third points lie in the interior of the ellipse. A covertex of the ellipse is at $(2\sqrt2,0)$, which makes the last point clearly exterior to the ellipse.

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  • $\begingroup$ Hi. The sum of distances from the foci to any point on the ellipse is equal to the length of the major axis. In this case, $2b$. I wonder how you have written it to be $2\sqrt2a$. $\endgroup$
    – aarbee
    May 7 '20 at 7:36
  • $\begingroup$ @Ramit The covertex and foci of this ellipse form an isosceles right triangle. $\endgroup$
    – amd
    May 7 '20 at 22:54

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