<s>[INST] 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?
 [/INST] HINT :
You found $m_1=-\dfrac{1}{2}$ and $m_2=2$. This is correct. Then
$$
\begin{align}
4x^2+y^2&=72\\
y^2&=72-4x^2\\
y&=\sqrt{72-4x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}(72-4x^2)^{\Large\frac12}\\
m_2&=\frac12(72-4x^2)^{-\Large\frac12}(-8x)\\
2&=-\frac{4x}{\sqrt{72-4x^2}}\\
\sqrt{72-4x^2}&=-2x\\
72-4x^2&=(-2x)^2\\
72-4x^2&=4x^2\\
x^2&=9\\
x&=\pm3
\end{align}
$$
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