A helicopter is flying along the curve given by y = x^2+7. A soldier placed at (3,7) wants to shoot down helicopter...

Question

A helicopter is flying along the curve given by y = x^2 + 7. A soldier placed at (3,7) wants to shoot down the helicopter. He would find it feasible to do so when It is closest to him. Find this closest distance at which he can shoot the helicopter.

Answer 1

sqrt(5)
the distance between the helicopter and the shooter will be minimum when the line joining the shooter and the parabola is perpendicular to the tangent at the point of contact. Since the tangent at any point to the curve is given by differential of the equation of parabola i.e. (dy/dx) (x^2+7) i.e. 2x hence the slope of the line joining the shooter and the parabola being 90 degrees to it has a slope of (-1/2x). This line is passing through location of the shooter at one end (3,7) The equation of this line is ((y-7)/(x-3))=(-0.5/x).
This line will cross the parabola at y=x^3+7. substitution gives 2x^3+x-3=0. The only real root of this equation is x=1. when x=1 on the parabola y=8. The distance between the shooter and the helicopter is distance between (1,8) and (3,7) which is sqrt(5).

Answer 2

x=1,y=8, after solving for the min distance between points 3,7 and (x,x^2+7).

Comments

we have to find the distance between (1,8) and (3,7) the two end points of the line. what you have written also comes to 3,7 and 1,8 at x=1, the solution abscissa.

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yeah rightly pointed out, distance was asked, not the location, distance should be √5 I.e. 2.24