JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 4)
Let $$\mathrm{P}$$ and $$\mathrm{Q}$$ 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 :
$$\left(0, \frac{1}{4}\right)$$
$$\left(\frac{1}{2}, \frac{3}{4}\right)$$
$$\left(\frac{1}{4}, \frac{1}{2}\right)$$
$$\left(\frac{3}{4}, 1\right)$$
Explanation
$$y = mx + 2a + {1 \over {{m^2}}}$$ (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 $$ \simeq {{ - 8} \over 5} = \tan \theta $$
$$ \Rightarrow \cos \theta \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta \simeq {8 \over {\sqrt {89} }}$$
$${x_p} = 1 + \cos \theta \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)$$
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