Question

find the point on the parabola y=x^2 that is closest to the point (0 2)

Answer 1

The answer I got was x=0,(sqrt6)/x, -(stqrt6)/2, So would the point be ((sqrt6)/x, -(stqrt6)/2)

Answer 2

at x=0
\[\frac{ d^2D }{ dx^2 }=0-6=-6<0\]
so D or d has a relative maxima at x=0
\[at~x=\pm \sqrt{\frac{ 3 }{ 2 }},\]
\[\frac{ d^2D }{ dx^2 }=12 \times \frac{ 3 }{ 2 }-6=12>0\]
so D or d has a relative minima at \[x=\pm \sqrt{\frac{ 3 }{ 2 }}\]
or \[x=\pm \frac{ \sqrt{6} }{ 2 },y=x^2=\frac{ 3 }{ 2 }\]
so nearest points are \[\left( -\frac{ \sqrt{6} }{ 2 },\frac{ 3 }{ 2 } \right)~and ~\left( \frac{ \sqrt{6} }{ 2 },\frac{ 3 }{ 2 } \right)\]