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Find the equation of the ellipse with the foci at (0,3) and (0, -3) for which the constant referred to in the definition is $6\sqrt{3}$

So I'm quite confused with this one, I know the answer is $3x^2+2y^2=54$ through trial and errror, but is there any way to solve it without trial and error?

The constant referred to is the sum of the distances from the foci right? How do I use it? I've been experimenting with how you can get $6 \sqrt3$ from the equation like using distance formula etc. I'm kinda stuck with this one, can someone please explain thanks!

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The sum of the distances to the foci is $6\sqrt{3}$. Via the "distance formula," this translates to $$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}.$$ Now let us grind. Rewrite as $$\sqrt{x^2+(y-3)^2}=6\sqrt{3}-\sqrt{x^2+(y+3)^2}.$$ Square both sides. We arrive at $$x^2+(y-3)^2=108 -12\sqrt{3}\sqrt{x^2+(y+3)^2} +x^2+(y+3)^2.$$ There is some cancellation. After the cancellation, we can divide through by $12$ and arrive at $$\sqrt{3}\sqrt{x^2+(y+3)^2}=9+y.$$ Square both sides again. Again, there is some cancellation, and we arrive at $$3x^2+2y^2=54.$$

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An ellipse is the figure consisting of all those points for which the sum of their distances to two fixed points (called the foci) is a constant$=2a$, where the lengths of the major & the minor axes are $a,b$ respectively.

So if $(h,k)$,be any point on the ellipse, $$\sqrt{(h-0)^2+(k-3)^2}+\sqrt{(h-0)^2+(k+3)^2}=6\sqrt3$$

On simplification we get, $$3h^2+2k^2=54$$

So, the equation of the ellipse is $$3x^2+2y^2=54$$


Alternatively,

as the foci $(0,\pm3)$ lie on the major axis, its equation will be $x=0$.

So, the $Y$ axis the major axis.

As the centre of the midpoint of the foci, hence centre will be $\left(\frac{0+0}2,\frac{-3+3}2\right)$ i.e., $(0,0)$, the origin.

According to the given problem

$2a=6\sqrt3\implies a=3\sqrt3$

and as the co-ordinate of the foci are $(0,\pm ae),$ $\implies ae=3\implies e=\frac1{\sqrt3}$

We know, $b^2=a^2(1-e^2)=27\left(1-\frac13\right)=18$

So, the equation of the ellipse is $$\frac{(x-0)^2}{18}+\frac{(y-0)^2}{27}=1$$

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Let $(x,y)$ be a point of ellipse then from definition $$\sqrt{(x-0)^2+(y-3)^2}+\sqrt{(x-0)^2+(y+3)^2}=6\sqrt{3}$$ $$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}$$ $$2x^2+2y^2+18+2\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=108$$ $$x^2+y^2+9+\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=54$$ $$(x^2+y^2-45)^2=(x^2+(y+3)^2)(x^2+(y-3)^2)$$ $$(x^2+y^2)^2-90x^2-90y^2+45^2=x^4+x^2(y+3)^2+x^2(y-3)^2+(y^2-9)^2$$ $$2x^2y^2+y^4-90x^2-90y^2+45^2=x^2(2y^2+18)+y^4-18y^2+81$$ $$-90x^2-90y^2+45^2=18x^2-18y^2+81$$ $$108y^2+72x^2=36\cdot54$$ $$3x^2+2y^2=54$$

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Note that the general form of the equation for an ellipse is $$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1.$$You are right that the constant they give is the sum of the distances from the foci, and note that it is $2a$ (why?). This means we have the equation $$\left(\frac{x}{3\sqrt{3}}\right)^2+\left(\frac{y}{b}\right)^2=1.$$ We can use the Pythagorean theorem to solve for $b$ since we know what the center of the ellipse is and we know the sum of the distances. Hence, solving for $b$ gives $$b=2\sqrt{(3-0)^2+(y-0)^2}=6\sqrt{3}=\sqrt{18}=3\sqrt{2}.$$Plugging these values in, we have $$\frac{x^2}{27}+\frac{y^2}{18}=1\Longrightarrow 2x^2+3y^2=54.$$

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