Shortest Distance Between Line y−x=1 and Curve x=y²

JEE Main 2011. Previous Year Question.

Problem

The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is:

Key insight. Since the parabola $x=y^2$ is easy to parametrise as $(t^2,t)$, the distance from any point on it to the line can be written as a single function of $t$ alone. Minimising that function — an ordinary one-variable calculus problem — gives the shortest distance directly, without needing calculus on the line itself.

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Approach

Write the line in standard form $x-y+1=0$, and parametrise a general point on the parabola as $(t^2,t)$. Use the point-to-line distance formula to express the distance as a function of $t$, then minimise this function using calculus (or by completing the square) to find the shortest distance.

Solution

Step 1 — Write the line in standard form

$$y-x=1 \implies x-y+1=0$$

Step 2 — Parametrise a point on the parabola

For $x=y^2$, let $y=t$, so $x=t^2$. A general point on the curve is $(t^2, t)$.

Step 3 — Write the distance from this point to the line

Using the point-to-line distance formula for $x-y+1=0$:

$$d(t) = \frac{|t^2-t+1|}{\sqrt{1^2+(-1)^2}} = \frac{|t^2-t+1|}{\sqrt2}$$

Step 4 — Check the sign of t² − t + 1

The discriminant of $t^2-t+1$ is $1-4=-3<0$, so this expression is always positive — the absolute value can be dropped:

$$d(t) = \frac{t^2-t+1}{\sqrt2}$$

Step 5 — Minimise t² − t + 1

Completing the square (or using calculus, $\frac{d}{dt}(t^2-t+1)=2t-1=0 \implies t=\frac{1}{2}$):

$$t^2-t+1 = \left(t-\frac{1}{2}\right)^2+\frac{3}{4}$$

This is minimised when $t=\dfrac{1}{2}$, giving a minimum value of $\dfrac{3}{4}$.

Step 6 — Compute the minimum distance

$$d_{\min} = \frac{\frac{3}{4}}{\sqrt2} = \frac{3}{4\sqrt2} = \frac{3\sqrt2}{8}$$

Answer

Common mistakes

• Forgetting to check whether $t^2-t+1$ can be negative before dropping the absolute value. Here the discriminant confirms it’s always positive, but this check is essential — skipping it could lead to incorrectly minimising the wrong (unsigned) expression.
• Minimising $t^2-t+1$ itself and forgetting to divide by $\sqrt2$ at the end. The minimum of the numerator alone ($\frac{3}{4}$) is not the final answer — it still needs to be divided by the line’s normalising factor $\sqrt{1^2+(-1)^2}=\sqrt2$.

Practise next

• Find the shortest distance between the line $x-y=2$ and the curve $y=x^2$, using the same parametrise-and-minimise approach.