# Find the points on the hyperbola x^2/4-y^2=1 that are closest to the point (5,0).

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### 1 Answer

Let the point on the hyperbola that is closest to (5,0) be (X, Y)

The distance between the two is `D = sqrt((X - 5)^2 + Y^2)`

The hyperbola given is: x^2/4 - y^2 = 1

=> y^2 = x^2/4 - 1

`D = sqrt((X - 5)^2 + X^2/4 - 1)`

=> `D = sqrt(X^2 + 25 - 10X + X^2/4 - 1)`

=> `D = sqrt((5*X^2/4) - 10X + 24)`

Now to minimize D, we have to solve D' = 0

D' = `((1/2)*(X*(5/2) - 10))/sqrt((3*X^2/4) - 10X + 24)`

D' = 0

=> X*(5/2) = 10

=> X = 20/5

=> X = 4

Y = sqrt(X^2/4 - 1)

=> Y = sqrt 3 and Y = -sqrt 3

**The required points are (4, sqrt 3) and (4, -sqrt 3)**