<s>[INST] equation of an ellipse given its center and two tangent lines There exists an ellipse centered at (0,0) with two tangent lines given by $y=-\frac{1}{2}x + \frac{\sqrt{39}}{2}$ and $y=\frac{1}{3}x + \frac{7}{3}$.  Find the ellipse.
So, I used the equation of one line and substituted it into the equation of an ellipse:
$\frac{x^2}{a^2} + \frac{(\frac{1}{3}x + \frac{7}{3})^2}{b^2}$=1.
Then I get a quadratic equation in $x$:
$(9b^2+a^2)x^2 + (14a^2)x + (49a^2 - 9a^2b^2)$ = 0
When I solve for $x$ using the quadratic formula, I get a very complicated expression in terms of $a$ and $b$:
$x = \frac{-14a^2 \pm \sqrt{196a^2 - 1764a^2b^2 + 324a^2b^4 - 98a^4 + 36a^4b^2}}{18b^2 + 2a^2} $
At this point, I figure that there must be a simpler way of solving this problem.  Otherwise, I have to plug this expression into the original equation of an ellipse to get an equation in terms of a and b.  Then, repeat the entire process with the second line to get a second equation in terms of a and b, and then use that system to solve for both a and b.
I've also thought about doing this with derivatives:
The slope of the second line is $\frac{7}{3}$, so the derivative at the $(x,y)$ coordinate where that line intersects the ellipse, of the ellipse, using implicit differentiation, yields (after solving):
$\frac{dy}{dx} = -\frac{xb^2}{ya^2} = \frac{7}{3}$
This, again, seems like it would lead to a lot of messy algebra if I were to plug this back into the original equation after solving for x:
$x=-\frac{7ya^2}{3b^2}$
Then plugging in:
$\frac{(-\frac{7ya^2}{3b^2})^2}{a^2} + \frac{y^2}{b^2} = 1$
As you can see, the algebra gets messy.  Is there a more simplistic way of going about this?
 [/INST] You can prove that the tangent line to an ellipse of equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
at a point $(s,t)$ on the ellipse is
$$
\frac{sx}{a^2}+\frac{ty}{b^2}=1
$$
You can rewrite the equations of your tangent lines as
$$
\frac{x}{\sqrt{39}}+\frac{2y}{\sqrt{39}}=1,
\qquad
\frac{-x}{7}+\frac{3y}{7}=1
$$
Thus there must exist $(s,t)$ and $(u,v)$ on the ellipse such that
$$
\frac{s}{a^2}=\frac{1}{\sqrt{39}},
\quad
\frac{t}{b^2}=\frac{2}{\sqrt{39}},
\qquad
\frac{u}{a^2}=-\frac{1}{7},
\quad
\frac{v}{b^2}=\frac{3}{7}
$$
With the assumption that $\dfrac{s^2}{a^2}+\dfrac{t^2}{b^2}=1$ we get
$$
\frac{s}{\sqrt{39}}+\frac{2t}{\sqrt{39}}=1
$$
and, similarly,
$$
-\frac{u}{7}+\frac{3v}{7}=1
$$
Now $\dfrac{s}{\sqrt{39}}=\dfrac{a^2}{39}$ and $\dfrac{2t}{\sqrt{39}}=\dfrac{4b^2}{39}$, so we get
$a^2+4b^2=39$.
Similarly $-\dfrac{u}{7}=\dfrac{a^2}{49}$ and $\dfrac{3v}{7}=\dfrac{9b^2}{49}$, so we get $a^2+9b^2=49$.
This implies $b^2=2$ and $a^2=31$, so your ellipse has equation
$$
\frac{x^2}{31}+\frac{y^2}{2}=1
$$
</s>