Question

The point on the curve y² = 4x which is nearest to, the point (2,1) is
(a) $1,2\sqrt{2}$

(b) (1,2)

(c) (1,$-$2)

(d) ($-$2,1)

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\left(1,2\right) \\ \mathrm{Let}\mathrm{the}\mathrm{required}\mathrm{point}\mathrm{be}\left(x,y\right).\mathrm{Then}, \\ {y}^2=4x \\ \Rightarrow x=\frac{y^2}{4}...\left(1\right) \\ \mathrm{Now}, \\ d=\sqrt{{\left(x-2\right)}^2+{\left(y-1\right)}^2} \\ \mathrm{S}\text{quaring both sides, we get} \\ \Rightarrow d^2={\left(x-2\right)}^2+{\left(y-1\right)}^2 \\ \Rightarrow d^2={\left(\frac{y^2}{4}-2\right)}^2+{\left(y-1\right)}^2 \\ \Rightarrow d^2=\frac{y^4}{16}+4-y^2+y^2+1-2y\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \mathrm{Now}, \\ Z=d^2=\frac{y^4}{16}+4-y^2+y^2+1-2y \\ \Rightarrow \frac{dZ}{dy}=\frac{y^3}{4}-2y+2y-2 \\ \Rightarrow \frac{dZ}{dy}=\frac{y^3}{4}-2 \\ \Rightarrow \frac{y^3}{4}-2=0 \\ \Rightarrow y^3=8 \\ \Rightarrow y=2 \\ \text{Substituting the value of}\mathit{\text{y}}\text{in}\left(\text{1}\right)\text{, we get} \\ x=1 \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{y}^2}=\frac{3y^2}{4} \\ \Rightarrow \frac{d^{2}Z}{d{y}^2}=\frac{3{\left(2\right)}^2}{4}=3>0 \\ \text{So, the nearest point is}\left(1,2\right). \end{gathered}$$