Question

Find the points on the curve xy 2 = 250 that are closest to the origin.

Answer 1

Hello,
All we have to do it to minimise
`f= √(x^2+y^2) = \sqrt{x^2+ (250)/(x^2) } \\ (df)/(dx) = (1)/(2)* \frac{2x- (2*250)/(x^3)}{ \sqrt{x^2+ (250)/(x^2) } }\\ \rightarrow x- (250)/(x^3) =0\\ \rightarrow x= \sqrt[4]{250} \approx 3.9763\\ \rightarrow y=250^{ (3)/(8) }\approx 7,9292\\ Sol=\{(\sqrt[4]{250}, \sqrt[8]{250^3} ), (\sqrt[4]{250}, -\sqrt[8]{250^3} )\} \\`

Answer 2

hello :
let : M(x,y) in the curve and the origin : O (0,0)
calculate : OA
OA² = x²+y².....(1)
but : xy² = 250
so : y² = 250/x
subsct in (1) :
OA² = x² + 250/x
let : f(x) = x² +250/x
calculate the minumum of : f
f'(x) = 2x - 250/x²
f'(x) = 0 : 250/x² = 2x
x^3 = 125 = 5^3
x= 5
f(5) =5² +250/3 =325/3....(The closest distance)

M(x,y) in the curve : x=5 : 5y² = 250
y² =50
y = √(50) or y = - √(50)
two points on the curve xy 2 = 250 that are closest to the origin:
M1 (5 , √(50)) , M2 (5 , -√(50))