AIEEE 2011 (Offline)Question 1
The shortest distance between the line $y-x=1$ and the curve $x=y^{2}$ is
Differentiation and its Application
Solution
Concept. Take a point $(t^2,t)$ on the parabola and minimise its distance from the line. Formulas used. (1) Distance of $(x_1,y_1)$ from $ax+by+c=0$: $\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$Step 1 - Distance. The line is $x-y+1=0$. By (1), $$d(t)=\frac{t^2-t+1}{\sqrt2}\quad(t^2-t+1>0\text{ always}).$$ Step 2 - Minimise. $t^2-t+1$ is least at $t=\frac12$, value $\frac34$. Step 3 - Value. $d=\dfrac{3}{4\sqrt2}=\dfrac{3\sqrt2}{8}$. Answer: (A)