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?

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 $y=2x+12$. 
