Q.The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is (A) $(2\sqrt{2}, 4)$ (B) $(2\sqrt{2}, 0)$ (C) $(0, 0)$ (D) $(2, 2)$

Concept: Distance Minimization — we minimise the squared distance to avoid the square root.

Let a general point on $x^2 = 2y$ be $(t, t^2/2)$. The squared distance to $(0,5)$ is

Differentiate:

Set to zero: $t = 0$ or $t = \pm 2\sqrt{2}$.

Check distances: …

The core idea here is distance minimization — but with a twist. You’re not just finding the shortest straight line from $(0,5)$ to any point; the point must lie on the curve $x^2 = 2y$. So we’re looking for the point on this parabola that is closest to $(0,5)$.

Why does minimizing the square of the distance work? Because the square root function is monotonic — the point that minimizes $d^2$ also minimizes $d$, and avoiding the square root makes differentiation much cleaner. This is a standard trick in optimization problems.

Let’s work through it.

1. Parametrize the curve.

   The parabola $x^2 = 2y$ can be written as $y = \frac{x^2}{2}$. So any point on it has coordinates $(t, \frac{t^2}{2})$, where $t$ is a real parameter (the $x$-coordinate).

2. Write the squared distance.

   Distance from $(t, \frac{t^2}{2})$ to $(0,5)$ is

Simplify:

3. Expand and simplify.

So we need to minimize $f(t) = \frac{t^4}{4} - 4t^2 + 25$.

4. Differentiate and find critical points.

Set $f'(t) = 0$:

5. Check which gives the minimum. Use the second derivative: $f''(t) = 3t^2 - 8$.
   • At $t = 0$: $f''(0) = -8 < 0$ → local maximum.
   • At $t = \pm 2\sqrt{2}$: $f''(2\sqrt{2}) = 3(8) - 8 = 16 > 0$ → local minimum. …

Method: Finding the Point on a Curve Nearest to a Fixed Point

This method applies to any "shortest distance from a point to a curve" question, whether the curve is given explicitly as $y=f(x)$ or, as here, in an implicit form like $x^2=2y$.

Steps

Step 1: Parametrize a general point on the curve using a single variable

Rewrite the curve so one coordinate is expressed in terms of the other, and let that free coordinate be your parameter. For $x^2=2y$, write $y=\dfrac{x^2}{2}$, so a general point is $(t,\,t^2/2)$ for parameter $t$.

Step 2: Write the squared distance to the fixed point as a function of that one variable

If the fixed point is $(a,b)$,

Always minimize the squared distance, not the distance itself — the square root only makes differentiation harder and does not change where the minimum occurs, since squaring is order-preserving for non-negative values.

Step 3: Expand, differentiate, and find every critical point

Expand $D(t)$ into a polynomial in $t$, differentiate, and solve $D'(t)=0$. A quartic-in-$t$ distance function like this one commonly has more than one critical point — do not stop at the first root. …

Common Mistakes

Mistake 1: Minimizing the distance $D(t)$ itself instead of the squared distance

Carrying the square root through differentiation multiplies the chance of an algebra slip. Since minimizing $D^2$ always gives the same optimal $t$ as minimizing $D$ (both are non-negative), there is no reason to keep the square root — drop it from the start.

Mistake 2: Accepting the first critical point found without comparing all candidates …