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I have to find the equation of an ellipse which passes through the point $(3, 2)$, has center at the origin and major axis along the y-axis, i.e., is a vertical ellipse. No other info is given. I've tried a lot but it seems too hard to figure out.

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  • $\begingroup$ So, do you know the general form of the equation of an ellipse with center at the origin and axes along the coordinate axes? $\endgroup$ – Gerry Myerson Mar 24 '14 at 12:38
  • $\begingroup$ @GerryMyerson : Yes. $\endgroup$ – Mayank Kumar Mar 24 '14 at 14:26
  • $\begingroup$ Good: what is it? $\endgroup$ – Gerry Myerson Mar 24 '14 at 23:32
  • $\begingroup$ @GerryMyerson: $\frac{(x^2)}{(a^2)}+\frac{(y^2)}{(b^2)}=1$ $\endgroup$ – Mayank Kumar Mar 25 '14 at 11:53
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Just put your point $(3,2)$ into $\frac {x^2}{a^2}+\frac{y^2}{b^2}=1$ and solve for $b$ to get $$ b=-\frac {2a}{\sqrt{a^2-9}}. $$ Set $a\gt 3$ and you get the family of ellipses that pass through $(3,2)$ and have center at the origin, since $$ \frac{3^2}{a^2}+\frac{2^2}{\left(-\frac {2a}{\sqrt{a^2-9}}\right)^2}=1 $$ is true for any $a$.

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    $\begingroup$ If $a\le3$, then $\sqrt{a^2-9}$ isn't real. $\endgroup$ – Gerry Myerson Mar 26 '14 at 9:48
  • $\begingroup$ @GerryMyerson yes sorry, I edited... $\endgroup$ – draks ... Mar 26 '14 at 11:16
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    $\begingroup$ If $a\ge b$, the major axis won't be along the $y$-axis, so there's a constraint on $a$. $\endgroup$ – Gerry Myerson Mar 26 '14 at 11:50

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