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Question Number 131188 by abdurehime last updated on 02/Feb/21

$$\mathrm{find}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)=1-{\mathrm{x}}^2$$$$\mathrm{that}\mathrm{are}\mathrm{closest}\mathrm{to}\mathrm{O}(0,0)$$

Answered by john_santu last updated on 02/Feb/21

$$letP(x,y)isthepointonthe$$$$curve.$$$$letthedistancebetweenpointPtopointO$$$$=d=\sqrt{x^2+y^2}=\sqrt{1-y+y^2}$$$$or{d}^2=y^2-y+1;y^2-y+1-d^2=0$$$$bytangency\mathrm{\Delta }=0$$$$\Rightarrow 1-4(1-d^2)=0;1-d^2=\frac{1}{4}$$$$\Rightarrow d^2=\frac{3}{4}\Rightarrow y^2-y+\frac{1}{4}=0$$$${(y-\frac{1}{2})}^2=0\to \{\begin{array}{l}y=\frac{1}{2}\\ x=\pm \frac{1}{2}\sqrt{2}\end{array}$$$$weget\to \{\begin{array}{l}{P}_1(\frac{1}{2}\sqrt{2},\frac{1}{2})\\ P_2(-\frac{1}{2}\sqrt{2},\frac{1}{2})\end{array}$$

Commented by abdurehime last updated on 02/Feb/21

$$\mathrm{i}\mathrm{satisfied}100\mathrm{\%}\mathrm{allh}\mathrm{bless}\mathrm{you}$$

Answered by mr W last updated on 02/Feb/21

$$saythepointis(x,y)with$$$$y=1-x^2$$$$thedistancefrom(x,y)to(0,0)isd.$$$$d^2=x^2+y^2=x^2+{(1-x^2)}^2$$$$=x^4-x^2+1$$$$={(x^2-\frac{1}{2})}^2+\frac{3}{4}⩾\frac{3}{4}$$$$\Rightarrow d_{min}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$$$$at{x}^2=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt{2}}{2},y=1-\frac{1}{2}=\frac{1}{2}$$$$\Rightarrow point(-\frac{\sqrt{2}}{2},\frac{1}{2})orpoint(\frac{\sqrt{2}}{2},\frac{1}{2})$$

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