Question 1082518
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An airplane is flying along the hyperbolic path illustrated in the figure. 
If an equation of the path is 2y^2-x^2=8 , determine how close the airplane comes to a town located at (3,0). 

(Hint: Let S denote the square of the distance from a point on the path to(3,0), and find the minimum value of S.)
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1.  Our curve is specific !  If (x,y) is the point on the curve then

    2y^2 - x^2 = 8,  which implies  y^2 = {{{(8+x^2)/2}}}.   (1)



2.  The distance from ANY point (x,y) on the coordinate plane to the point (3,0) is

    {{{d^2}}} = {{{(x-3)^2 + y^2}}}.    (2)

    Now, if the point lies on the hyperbola, you have (1), and you can substitute this expression for {{{y^2}}} into the formula (2).

    You will get

    {{{d^2}}} = {{{(x-3)^2}}} + {{{(8+x^2)/2}}} = {{{x^2 - 6x + 9 + 4}}} + {{{x^2/2}}} = {{{(3/2)*x^2 - 6x + 13}}}.   (3)

    So, you need to minimize (3). In other words, you need to find the value of "x" which minimizes this quadratic function.



4.  The quadratic function of the general form q(x) = {{{ax^2 + bx + c}}} achieves the maximum at x = {{{-b/(2a)}}}.

    In your case this "x" is x = - {{{((-6))/(2*(3/2))}}} = {{{6/3}}} = 2.


    So, we found the value of "x". It is x= 2.

    Then the corresponding value of "y" on your curve is 

         {{{2y^2 - 2^2}}} = 8  ---->  {{{2y^2}}} = 8 + 4 = 12  ---->  {{{y^2}}} = {{{12/2}}} = 6.

     Thus your "closest" point on the curve is  (x,y) = ({{{2}}},{{{sqrt(6)}}}).



5.  Now you can find that minimal distance:

    {{{d^2}}} = {{{(x-3)^2 + y^2}}} = {{{(2-3)^2 + 6}}} = 1 + 6 = 7.


    Hence, the minimal distance itself is  {{{sqrt(7)}}} = 2.646 (approximately).
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My congratulations !  &nbsp;&nbsp;The problem is solved.