Equation of tangent perpendicular to line I've got the homework question which I cannot solve.

Find the equation of the tangents to $4x^2+y^2=72$ that are
  perpendicular to the line $2y+x+3=0$.

What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation perpendicular to the tangent of $4x^2+y^2=72$ is $m_2 = 2$, since $m_1m_2 = -1$.
I've found the derivative of $4x^2+y^2=72$ which is $\dfrac{dy}{dx} = -\dfrac{4x}{y}$.
The next thing I have done is set $m_2=\dfrac{dy}{dx}$.
Now what?
 A: $$2y+x+3=0$$
The important thing is that you know what being perpendicular means. In terms of slopes, one line has to have a slope that's equal to the negative reciprocal of the other. This is a property you should know from your Algebra class. For this problem, we want to find the slope of the line that's perpendicular to the line we have above. To do that, we will first need to find the slope of the line above. Let's rearrange stuff to find it:
$$2y=-x-3 \\ y=-\frac12x-1.5$$
You can get $\frac{dy}{dx}$ of this, but really it's in the form of $y=mx+b$ and we know that the slope of this line is $-\frac12$. Thus, the slope of the line perpendicular to this one is $2$.
$$4x^2+y^2=72$$
We've got the slope, but not the line yet. We want to find a line that will be tangent to the ellipse that was given. How do we find that? Start with finding the point at which it is tangent. You probably already know how to do this. First take the derivative. For this problem, implicit differentiation is not a really good idea. Thus, isolate for $y$ first and then solve it:
$$y=\sqrt{72-4x^2}$$
$$\frac{dy}{dx}=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
Now that you have the derivative, find the value for $x$ where the slope of the ellipse is the same as the perpendicular line. In other words, $\frac{dy}{dx}=2$:
$$2=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
A little messy, but you'll find that $x=\pm3$. Therefore the point on the ellipse where the tangent will have this slope is $(\pm3, \pm6)$ (plugged in to the original to get the $y-$value). From this point, you can easily get the line now. You've got $y=mx+b$. $m$ is $2$, and now to find $b$ we'll plug in $x/y$:
$$6=(2)(-3)+b$$
$$b=12$$
Or for $x=3$, $b=-12$.
Our final answer ends up being:
$$\therefore y=2x\pm12$$
Pictorial proof:

A: HINT :
You found $m_1=-\dfrac{1}{2}$ and $m_2=2$. This is correct. Then
$$
\begin{align}
4x^2+y^2&=72\\
y^2&=72-4x^2\\
y&=\sqrt{72-4x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}(72-4x^2)^{\Large\frac12}\\
m_2&=\frac12(72-4x^2)^{-\Large\frac12}(-8x)\\
2&=-\frac{4x}{\sqrt{72-4x^2}}\\
\sqrt{72-4x^2}&=-2x\\
72-4x^2&=(-2x)^2\\
72-4x^2&=4x^2\\
x^2&=9\\
x&=\pm3
\end{align}
$$
A: Let the point of tangency be the point  $(x,y)$ on our curve. Your calculation shows that $\frac{-4x}{y}=2$, that is, $y=-2x$. 
Also, we have $4x^2+y^2=72$. Substituting $-2x$ for $y$ in this equation, we get $8x^2=72$, and therefore $x=\pm 3$. That gives the two points $(3,-6)$ and $(-3,6)$.
The line with slope $2$ through $(3,-6)$ has equation $y-(-6)=2(x-3)$, which simplifies to $y=2x-12$. The other line has  equation $y=2x+12$. 
