Answer
The graph must be an ellipse with endpoints (±2,0) and $(\pm 1,0),$ the point $(-2,0),$ and the vector $\nabla f$ pointing out from the point.
Work Step by Step
To get its corresponding level curve, we begin by replacing point $P$ in $f(x, y)$
\[
\begin{array}{c}
4=(-2)^{2}+4(0)^{2}=f(-2,0) \\
x^{2}+4 y^{2}=4
\end{array}
\]
To calculate the gradient, we invoke the gradient formula $f_{x} \mathbf{i}+f_{y} \mathbf{j}=\nabla f$ .
\[
\begin{array}{c}
f_{x}=2 x=2(-2)=4 \text { and } f_{y}=8 x=8(0)=0 \\
\nabla f=-4 \mathbf{i}
\end{array}
\]
The graph must be an ellipse with endpoints (±2,0) and $(\pm 1,0),$ the point $(-2,0),$ and the vector $\nabla f$ pointing out from the point.
