Q.Find the shortest distance of the point $(0, c)$ from the parabola $y = x^2$, where $\dfrac{1}{2} \le c \le 5$.

Concept: Distance from a point to a curve — minimise the squared distance using calculus.

Let a general point on the parabola be $(t, t^2)$. The squared distance from $(0, c)$ is

Differentiate with respect to $t$ and set to zero:

So either $t = 0$ or $t^2 = c - \frac12$.

Since $\frac12 \le c \le 5$, the value $c - \frac12$ is non-negative, so $t^2 = c - \frac12$ is valid.

• For $t = 0$: distance $= |c| = c$ (since $c>0$). …

Squared distance. A general point on $y=x^2$ is $(t,t^2)$. Let

Critical points.

For $c\ge\tfrac12$ the second option is real.

Compare the values.

Their difference is

$c^2-\left(c-\tfrac14\right)=\left(c-\tfrac12\right)^2\ge 0,$ …

Method: Minimizing the Distance From a Point to a Curve

The general technique for "closest point on a curve" problems — and a reminder to check every critical point the algebra produces, not just the first one found.

Steps

Step 1: Parametrize a general point on the curve

Write a typical point on the curve using one parameter (here, a point on $y=x^2$ can be written $(t, t^2)$), then form the squared distance to the fixed point.

Step 2: Minimise the squared distance, not the distance itself

Since squaring preserves order for non-negative values, the same $t$ minimises both $D$ and $D^2$ — but $D^2$ avoids differentiating a square root.

Step 3: Differentiate, solve for ALL critical points, and check validity

This can factor to give more than one critical value of $t$ (or, as here, a condition on the fixed point's own parameter). Discard any critical value that falls outside the problem's stated range. …

Common Mistakes

Mistake 1: Stopping at the critical point $t=0$ and reporting distance $=c$

Why it's wrong: solving $\frac{d(D^2)}{dt}=0$ gives $2t\big[1+2(t^2-c)\big]=0$, which has two families of solutions — $t=0$ and $t^2=c-\tfrac12$ — but a student who only factors out $t$ and drops the bracket entirely gets just the first, weaker candidate. Correct approach: solve the full factored equation and keep every branch; here the second branch gives the genuinely smaller squared distance $c-\tfrac14$ for $c>\tfrac12$.

Mistake 2: Forgetting to check that $t^2=c-\tfrac12$ is a valid (real, in-range) solution …