Asked by demi
Find the x–coordinate of the point on the parabola y=x^1/2 nearest to the point (4, 0).
Answers
Answered by
Reiny
notice the point (4,0) does not lie ON the curve.
So your sketch should show that.
let that point be (a,b)
so we know b = √a or <b>a = b^2</b>
dy/dx = (1/2)x^(-1/2) or 1/(2√x)
so at the point (a,b) the slope of the tangent is 1/(2√a)
I will use the property that at closest point the tangent would be perpendicular to the line joining (a,b) to (4,0)
slope of line from (a,b) to (4,0) = b/(a-4)
then 1/(2√a) = (a-4)/-b
2√a(a-4) = -b
but remember that b = √a , so
2√a(a-4) = -√a
2(a-4) = -1
2a - 8 = -1
a = 7/2 , then b = √7/√2 or √14,2 after rationalizing it
the closest point is (7/2 , √14/2 )
So your sketch should show that.
let that point be (a,b)
so we know b = √a or <b>a = b^2</b>
dy/dx = (1/2)x^(-1/2) or 1/(2√x)
so at the point (a,b) the slope of the tangent is 1/(2√a)
I will use the property that at closest point the tangent would be perpendicular to the line joining (a,b) to (4,0)
slope of line from (a,b) to (4,0) = b/(a-4)
then 1/(2√a) = (a-4)/-b
2√a(a-4) = -b
but remember that b = √a , so
2√a(a-4) = -√a
2(a-4) = -1
2a - 8 = -1
a = 7/2 , then b = √7/√2 or √14,2 after rationalizing it
the closest point is (7/2 , √14/2 )
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