$F(x, y) = x^2 + 2y^2 − 6x + 4y − 7$
a. Find a differential equation whose solutions are the level curves of $F$.
b. Parametrize such level curves in order to prove that the gradient of $F$ is orthogonal to them.
c. What kind of curves are the solutions of the equation exhibited in (a)?
d. Is there a point of the plane for which there no solution of this equation passing across it? Justify.
The level curves of your function $z = F(x,y)$ are two dimensional curves you get by setting $z = k$, where $k$ is any number. Then, to find the differential equation of a., you must derivate the expression $F(x,y) = k$ with respect to $x$, for example. In your case, you will obtain: $2x + 4yy' - 6 + 4y' = 0$.
Gradient of $F$: $\vec{\nabla}F = (\partial F/ \partial x, \partial F / \partial y) = (2x - 6, 4y + 4)$.
If you rewrite the expression $F(x,y) = k$ you get \begin{equation} \frac{(x - 3)^2}{2} + (y+1)^2 = \frac{k + 18}{2} \iff \left(\frac{x-3}{\sqrt{k+18}}\right)^2 + \left(\frac{y+1}{\sqrt{\frac{k+18}{2}}}\right)^2 = 1, \end{equation} so a nice parametrization could be $(x(\theta),y(\theta)) = \left(3 + \sqrt{2}C\cos\theta,-1 + C\sin\theta\right)$, where $C = \sqrt{(k+18)/2}$. Then, you can see the orthogonality with the scalar product: \begin{equation} \vec{\nabla}F\;\cdot\;(x'(\theta),y'(\theta)) = (2\sqrt{2}C\cos\theta,4C\sin\theta)\;\cdot\;(-\sqrt{2}C\;·\sin\theta,C\cos\theta) = 0 \end{equation}
When I wrote $\frac{(x - 3)^2}{2} + (y+1)^2 = \frac{k + 18}{2}$, we can observe that the level curves are ellipses and, as the LHS is positive, $k$ must be greater or equal than $-18$.
Besides, as taking all the possible values of $k$, from $-18$ to $\infty$ you will be able to fill all the plane (notice that changing $k$ we are modifying the radius of the ellipse, so we can obtain any point of the plane).