Let P and Q be any points on the curves (x-1)^2+(y+1)^2=1 and y=x^2, respectively.

Question

Let $P$ and $Q$ be any points on the curves $(x-1)^2+(y+1)^2=1$ and $y=x^2$, respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval

Accepted Answer

$$y = mx + 2a + {1 \over {{m^2}}}$$ (Equation of normal to ${x^2} = 4ay$ in slope form) through $(1, - 1)$.

$4{m^3} + 6{m^2} + 1 = 0$

$$ \Rightarrow m \simeq  - 1.6$$

Slope of normal $$ \simeq {{ - 8} \over 5} = \tan \theta $$

$$ \Rightarrow \cos \theta  \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta  \simeq {8 \over {\sqrt {89} }}$$

$${x_p} = 1 + \cos \theta  \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)$$

Solutions · 1