Question

My question is hat if an Air force combat training jet execute, a high-speed maneuver along the a path 2ysquare minus x square that that is equal to 8. How close does the jet come to a town located at (3,0)? Let m denote the square of the distance from a point (x,y) on the path to (3,0) and finding the minimum value of m. What is square root m?

Answer 1

Answer:

Step-by-step explanation:

Let the function be:

f(x, y) = 2(y^2) - x^2 = 8

Since the shortest distance is basically a straight line from point (3, 0) to

f(3, y).

2y^2 - 3^2 = 8

2y^2 = 17

y = sqrt(17/2)

So the nearest point on the planes course to the point (3,0) is (3, sqrt(17/2))

Now lets use the distance formula:

x1 = 3

y1 = 0

x2 = 3

y2 = sqrt(17/2)

$=\sqrt{\left(3-3\right)^2+\left(\sqrt{\frac{17}{2}}-0\right)^2}$

$=\sqrt{(0)^2+(\sqrt{\frac{17}{2}})^2}$

$=\sqrt{(\sqrt{\frac{17}{2}})^2}$

$=\sqrt{\frac{17}{2} }$

So m in this case is just 17/2 and the square root of m is sqrt(17/2) = 2.91547594742