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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)

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Solution

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

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