Help with distance formula etc.

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(Calculus): on the curve y=sqrt(4x), what is the point that is closest to (5,0) (distance formula)and the function equation to represent it?

 

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(Calculus): on the curve y=sqrt(4x), what is the point that is closest to (5,0) (distance formula)and the function equation to represent it?

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Answer 1

y=√(4x)=2√x=2x^½, y'=1/√x=slope of tangent.

Let's say the point P(a,2√a) is the nearest point on the curve to Q(5,0), so the tangent at that point is 1/√a.

The equation of a circle centre Q is (x-5)²+y²=r², where r is the radius such that the tangent of the circle is also tangential to the curve.

2(x-5)+2yy'=0, y'=(5-x)/y=(5-x)/(2√x)=5/(2√x)-½√x. When x=a, y'=5/(2√a)-½√a.

But y'=1/√a when x=a and these two slopes have the same value:

1/√a=5/(2√a)-½√a, 2=5-a, so a=3. Therefore P(3,2√3) is where the tangents coincide. Length of r=PQ=√((5-3)²+(2√3-0)²)=√16=4.

GENERAL SOLUTION FOR ANY CURVE y=f(x) AND A GIVEN POINT Q(b,c)

y=f(x)), y'=f'(x)=slope of tangent.

Let's say the point P(a,f(a)) is the nearest point on the curve to Q(b,c), so the tangent on the curve at that point is f'(a).

The equation of a circle centre Q is (x-b)²+(y-c)²=r², where r is the radius such that the tangent of the circle is also tangential to the curve.

2(x-b)+2(y-c)y'=0, y'=(b-x)/(y-c)=(b-x)/(f(x)-c). When x=a, y'=(b-a)/(f(a)-c).

But y'=f'(a) when x=a and these two slopes have the same value:

f'(a)=(b-a)/(f(a)-c). Solve this for a in terms of b and c to find P(a,f(a)), which is where the tangents coincide. PQ=√((b-a)²+(c-f(a))²).

Applying these formulae to the given question:

f'(a)=1/√a, f(a)=√(4x), b=5, c=0⇒

1/√a=(5-a)/√(4a), 1=(5-a)/2, 2=5-a, a=3, f(a)=2√3, hence P(3,2√3). PQ=√(4+12)=4.