11. Consider the function $f(x, y)=x^{2}-4 y^{2}-8$ and the point $(-2,2)$. a. Find the unit vectors that give the direction of steepest ascent and steepest descent at $P$. b. Find a vector that points in a direction of no change in the function at $P$. a. What is the unit vector in the direction of steepest ascent at P? (Type exact answers, using radicals as needed.) What is the unit vector in the direction of steepest descent at P? (Type exact answers, using radicals as needed.) b. Which of the following vectors is in a direction of no change of the function at P? A. $\langle 4,-16\rangle$ B. $\langle 16,4\rangle$ C. $\langle 16,-4\rangle$ D. $\langle-4,16\rangle$

Step 1 ：First, we need to find the gradient of the function \(f(x, y)=x^{2}-4 y^{2}-8\), which is a vector of the partial derivatives of the function with respect to each variable. The gradient is \(\langle 2x, -8y \rangle\).

Step 2 ：Next, we substitute the point \((-2,2)\) into the gradient, which gives us \(\langle -4, -16 \rangle\).

Step 3 ：We then find the magnitude of the gradient, which is \(16.492422502470642\).

Step 4 ：The unit vector in the direction of steepest ascent is the gradient divided by its magnitude, which gives us \(\boxed{-0.24253563, -0.9701425}\).

Step 5 ：The unit vector in the direction of steepest descent is the negative of the unit vector in the direction of steepest ascent, which gives us \(\boxed{0.24253563, 0.9701425}\).

Step 6 ：Finally, we find a vector that points in a direction of no change in the function at \(P\). This is a vector that is orthogonal to the gradient at \(P\). We can find this by taking the dot product of the gradient and the vector and setting it equal to zero. The vectors we have are \(\langle 4, -16 \rangle\), \(\langle 16, 4 \rangle\), \(\langle 16, -4 \rangle\), and \(\langle -4, 16 \rangle\).

Step 7 ：After calculating, we find that the vector in a direction of no change in the function at \(P\) is \(\boxed{16, 4}\).