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An ellipse touches the $x$ axis. The length of the major axis is $2a$ while the minor axis is given as $2b$, What would be the locus of its focus? ($a>b$.)

My Approach:

If the foci are $(x_1,y_1$ and $(x_2,y_2)$, then $(x_1-x_2)^2 + (y_1 - y_2)^2 = 4(a^2 - b^2),$ also $y_1y_2 = b^2,$ since product of perpendicular subtended from the foci on any tangent is $b^2$.

These are the properties which seem relevant in this question but I am not able to use them to find out the answer. I also might be missing out on other things.

Any hints/solutions to the problem are appreciated.

amWhy
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marks_404
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  • In the standard equation of the ellipse, $a \gt b$ and $b^2 = a^2(1-e^2)$. So the relation b/w $a$ and $b$ is $a^2-b^2=a^2e^2$ where $e$ is the eccentricity and $0 < e <1$ Try these! – p_square May 28 '21 at 13:07
  • @Algebrology This is what I used to obtain $(x_1-x_2)^2 + (y_1 - y_2)^2 = 4(a^2 - b^2)$ is there anything else I can do with the property ? – marks_404 May 28 '21 at 13:15
  • Can you find the foci of this ellipse? – p_square May 28 '21 at 13:17
  • Not sure, getting confused because of x1,x2 which are both variable – marks_404 May 28 '21 at 13:32
  • The foci of the ellipse are $F_1(ae,0)$ and $F_2(-ae,0)$. Now I guess you can find the locus of the focus – p_square May 28 '21 at 14:48
  • @Algebrology How can you assume the major axis of the ellipse to be parallel to the x-axis? I don't think that should be right – marks_404 May 28 '21 at 15:25
  • Oh didn't read the question. I thought it was parallel to the x-axis – p_square May 28 '21 at 16:02
  • "Touches the $x$ axis" is too vague. Moreover, an ellipse has two foci. Are you sure the text of the question is correct? – Intelligenti pauca May 28 '21 at 18:36
  • From touches the x axis I infer that x axis is tangent to the ellipse. Yes the focii are two and its locii need's to be found. Language is the same as in the question – marks_404 May 28 '21 at 19:25
  • A focus can be at any position $(,)$ in the plane with the constraint $−≤≤+$ or $-−≤≤-+$, where $=\sqrt{^2−^2}$. – Intelligenti pauca May 28 '21 at 20:34
  • @Intelligentipauca now that you've mentioned this I see it too. It is very vague then. The locus in the answer was given as $(xy^2 + b^2x)^2 + (y^2 - b^2)^2y^2 = 4(a^2 - b^2)y^4$ does it hind to some extra info which they failed to mention in the question? – marks_404 May 28 '21 at 21:08
  • Yes, that curve is the result if the ellipse touches the $x$ axis at $(0,0)$. – Intelligenti pauca May 28 '21 at 22:11

1 Answers1

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HINT.

The ellipse touches the $x$ axis at the origin (see comments to the question). Hence, if $F_1=(x,y)$, then $\displaystyle F_2=\left(-{b^2\over y^2}x, {b^2\over y}\right)$.

Insert these into $\overline{F_1F_2}^2=4(a^2-b^2)$ to get the desired relation.

enter image description here

Intelligenti pauca
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