Question

The point on the curve 2(y)=x^2 nearest (4,1),is the point?

Answer 1

The nearest point is `(2,2)`

Explanation:

The parametric form of equation for curve `2y=x^2` is `(2t,2t^2)`

Now to find the point on the curve nearest to given point `(4,1)`, we should identify a point on the curve at which normal to the curve passes through `(4,1)`

To find equation of normal let us work out the slope of tangent, which is given by `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(4t)/2=2t`.

Hence slope of normal, which is perpendicular to tangent is `-1/(2t)` and as it passes through `(2t,2t^2)` is

`y-2t^2=-1/(2t)(x-2t)` andas it passes through `(4,1)`

`1-2t^2=-1/(2t)(4-2t)`

or `2t-4t^3=-4+2t`

or `4t^3-4=0` i.e. `t^3-1=0` or `(t-1)(t^2+t+1)=0`

as only real solution is `t=1`, the point on the curve is `(2,2)`

graph{(2y-x^2)((x-2)^2+(y-2)^2-0.03)((x-4)^2+(y-1)^2-0.03)(x+2y-6)=0 [-8.75, 11.25, -3.92, 6.08]}