UPSC 2018 Maths Optional Paper 1 Q2b — Step-by-Step Solution

13 marks · Section A

Maxima and minima of single-variable functions · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Find the shortest distance from the point $(1,0)$ to the parabola $y^2=4x$ .

Parametrise as $(t^2,2t)$ ; squared distance collapses to the perfect square $(t^2+1)^2$ , minimised at $t=0$ .

Points of $y^2=4x$ are $\big(t^2,\,2t\big)$ for $t\in\mathbb R$ (since $(2t)^2=4t^2$ ). The squared distance from $(1,0)$ to such a point is

D(t)=(t^2-1)^2+(2t-0)^2=t^4-2t^2+1+4t^2=t^4+2t^2+1=(t^2+1)^2.

$D(t)=(t^2+1)^2$ is minimised when $t^2+1$ is smallest, i.e. at $t=0$ , where $t^2+1=1$ . Then

D_{\min}=1\quad\Rightarrow\quad \text{distance}=\sqrt{D_{\min}}=1.

The nearest point is $(t^2,2t)=(0,0)$ , the vertex.

\boxed{\;\text{Shortest distance}=1,\ \text{attained at the vertex }(0,0).\;}