+81 What is the smallest distance between the origin and a point on the graph of y=1/sqrt(2) - (x^2-3)?
+118702 y=1√2−(x2−3)y=−x2+3+1√2y=−x2+6+√22
This is a concave down parabola with y intercept 6+√22≈3.707
And the y axis is the axis of symmetry.
A general point on this parabola is (x,[−x2+6+√22])
The distance square from (0,0) to the parabola for any given x value will be
d2=D=(x−0)2+(−x2+6+√22−0)2D=x2+(−x2+6+√22)2D=x2+x4+(−2∗6+√22)x2+(6+√22)2D=x2+x4+(−6−√2)x2+(36+2+12√24)D=x4+(1−6−√2)x2+(192+3√2)D=x4+(−5−√2)x2+(192+3√2) D′=4x3+2(−5−√2)xD will be minimum when D′=00=4x3+2(−5−√2)xI know just from thinking about the graph that x≠00=4x2+2(−5−√2)0=4x2−10−2√24x2=10+2√22x2=5+√2x2=5+√22x=±√5+√22
So the distance will be minimum when x=±√5+√22
Sub to get y
y=−[±√5+√22]2+6+√22y=−[5+√22]+6+√22y=−5−√22+6+√22y=12
So the closest points on the parabola to (0,0) are (√5+√22,12)and(−√5+√22,12)
So the distance from either both these points to the origin will be
d=√(5+√22)+14d=√(10+2√2+14)d=√(11+2√24)smallestdistance=√11+2√22smallestdistance≈1.86units
Here is the related graph:
LaTex:
y=\frac{1}{\sqrt2}-(x^2-3)\\
y=-x^2+3+\frac{1}{\sqrt2}\\
y=-x^2+\frac{6+\sqrt2}{2}\\
d^2=D=(x-0)^2+(-x^2+\frac{6+\sqrt2}{2}-0)^2\\
D=x^2+(-x^2+\frac{6+\sqrt2}{2})^2\\
D=x^2+x^4+(-2*\frac{6+\sqrt2}{2})x^2+(\frac{6+\sqrt2}{2})^2\\
D=x^2+x^4+(-6-\sqrt2)x^2+(\frac{36+2+12\sqrt2}{4})\\
D=x^4+(1-6-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\
D=x^4+(-5-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\~\\
D'=4x^3+2(-5-\sqrt2)x\\
\text{D will be minimum when }D'=0\\
0=4x^3+2(-5-\sqrt2)x\\
\text{I know just from thinking about the graph that }\;x\ne0\\
0=4x^2+2(-5-\sqrt2)\\
0=4x^2-10-2\sqrt2\\
4x^2=10+2\sqrt2\\
2x^2=5+\sqrt2\\
x^2=\frac{5+\sqrt2}{2}\\
x=\pm\sqrt{\frac{5+\sqrt2}{2}}\\
y=-\left[\pm\sqrt{\frac{5+\sqrt2}{2}}\right]^2+\frac{6+\sqrt2}{2}\\
y=-\left[\frac{5+\sqrt2}{2}\right]+\frac{6+\sqrt2}{2}\\
y=\frac{-5-\sqrt2}{2}+\frac{6+\sqrt2}{2}\\
y=\frac{1}{2}
\left(\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)\quad and \quad \left(-\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)
d=\sqrt{ \left(\frac{5+\sqrt2}{2}\right) + \frac{1}{4} }\\
d=\sqrt{ \left(\frac{10+2\sqrt2+1}{4}\right) }\\
d=\sqrt{ \left(\frac{11+2\sqrt2}{4}\right) }\\
smallest\;\; distance=\frac{\sqrt{ 11+2\sqrt2 }}{2}\\
smallest\;\; distance\approx 1.86\;units
