Question: For the ellipse $4x^2 + 9y^2 = 36$ and the function $2x + 3y = k$ find the values of $k$ that are a tangent to the ellipse.

I squared the function and got $ 4x^2 + 9y^2 = k^2$ 

When I solve it I get $k = \pm\sqrt 6$ don't know where I went wrong, as the answer key says it's $\pm2\sqrt6$ .

Any help for why I am missing the "$2$" at the end of the equation would be appreciated. Thank you.

Answer: 
$u = 2x, v = 3y\\
u^2 + v^2 = 36\\
u + v = k$

Now we have a circle and not an ellipse, and circles are easier to work with.

the tangent to the circle is perpendicular to the radius.
the point of tangency is then on the line $u = v$

$u,v = \pm 6\frac {\sqrt{2}}{2}\\
k = \pm 6{\sqrt{2}}$

Now that doesn't fit your answer key, either.
$$(a+b)^2=a^2+b^2+2ab$$

As [Jonathan Richard Lombardy](https://math.stackexchange.com/users/345851/jonathan-richard-lombardy) already stated this is what you're missing.