The point on the curve y^2=2x , which is nearest to the point (1,0) i…

The correct answer is Option (4) → (0,0)

Let D be the square of the distance from (1, 0) to a point (x, y) of the curve; D is least at the same point as the distance itself. Since \( y^2=2x \):

    \( \displaystyle D=(x-1)^2+y^2 \)

    \( \displaystyle =(x-1)^2+2x \)

    \( \displaystyle =x^2+1 \)

Since \( y^2=2x\ge 0 \), points of the curve have x ≥ 0.

    \( \displaystyle \frac{dD}{dx}=2x=0\ \Rightarrow\ x=0 \)

    \( \displaystyle \frac{d^2D}{dx^2}=2\gt 0 \)

So D is least at x = 0, where \( y^2=0 \), that is y = 0. The nearest point is (0, 0), at distance 1.

Shortcut: The nearest point must lie on the curve. Check each option in \( y^2=2x \):

    \( \displaystyle (2,0):\ 0\ne 4 \)

    \( \displaystyle (2\sqrt{2},4):\ 16\ne 4\sqrt{2} \)

    \( \displaystyle (0,2):\ 4\ne 0 \)
    \( \displaystyle (0,0):\ 0=0 \)

Only (0, 0) lies on the curve.