Minimum Distance
Geometry
Level
2
Find the minimum distance of the point P(0,2) from the curve 16x^2 -9y^2+36y-61=0
The answer is 1.25.
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1 solution
Let Q(a,b) be a point on the given curve from which we get the minimum value of PQ=[(a^2+(b-2)^2)]^1/2,......(i) by using the distance formula between two points. Since, Q is a point on the given curve, it should satify the given equation, which when simplified becomes 16a^2- 9(b-2)^2=25.....(ii)
Putting the value of a^2 from (ii) in (i), we get PQ= [{25 + 25(b-2)^2}/16]^1/2 , which will be minimum when (b-2) =0, hence the answer will be 5/4=1.25