New Notes Blog: Find the point on the parabola x+y^2=0 that is closest to the point (0,-3)

Thursday, 6 August 2015

Find the point on the parabola x+y^2=0 that is closest to the point (0,-3)

Let us say the coordinates of the point which is closest to (0,-3) is (a,b).

The distance(L) between these two points can be given as;

$\displaystyle{L}=\sqrt{{{{\left({0}-{a}\right)}}^{{2}}+{{\left(-{3}-{b}\right)}}^{{2}}}}$

By getting $\displaystyle{{L}}^{{2}}$

$\displaystyle{{L}}^{{2}}={{a}}^{{2}}+{{\left({3}+{b}\right)}}^{{2}}$ -----(1)

However (a,b) is on the parabola. So we can say;

$\displaystyle{a}+{{b}}^{{2}}={0}$

$\displaystyle{a}=-{{b}}^{{2}}$

By substituting $\displaystyle{a}=-{{b}}^{{2}}$ on equation (1)

$\displaystyle{{L}}^{{2}}={{b}}^{{4}}+{{\left({3}+{b}\right)}}^{{2}}$

$\displaystyle{{L}}^{{2}}={{b}}^{{4}}+{{b}}^{{2}}+{6}{b}+{9}$ ----(2)

The maximum/minimum of L is given...

Let us say the coordinates of the point which is closest to (0,-3) is (a,b).

The distance(L) between these two points can be given as;

$\displaystyle{L}=\sqrt{{{{\left({0}-{a}\right)}}^{{2}}+{{\left(-{3}-{b}\right)}}^{{2}}}}$

By getting $\displaystyle{{L}}^{{2}}$

$\displaystyle{{L}}^{{2}}={{a}}^{{2}}+{{\left({3}+{b}\right)}}^{{2}}$ -----(1)

However (a,b) is on the parabola. So we can say;

$\displaystyle{a}+{{b}}^{{2}}={0}$

$\displaystyle{a}=-{{b}}^{{2}}$

By substituting $\displaystyle{a}=-{{b}}^{{2}}$ on equation (1)

$\displaystyle{{L}}^{{2}}={{b}}^{{4}}+{{\left({3}+{b}\right)}}^{{2}}$

$\displaystyle{{L}}^{{2}}={{b}}^{{4}}+{{b}}^{{2}}+{6}{b}+{9}$ ----(2)

The maximum/minimum of L is given when dL/db = 0

By first derivative on (2)

2LdL/db = 4b^3+2b+6

For maximum and minimum dL/db = 0

$\displaystyle{4}{{b}}^{{3}}+{2}{b}+{6}={0}$

In these complex cases it is better to apply b = -1, b = +1 or b = 0 and see whether it solves equation.

You can see at b = -1 gives you one solution.

So we can write;

$\displaystyle{\left({b}+{1}\right)}{\left({p}{{b}}^{{2}}+{q}{b}+{r}\right)}={4}{{b}}^{{3}}+{2}{b}+{6}$ where b^2 is not equal to 0.

$\displaystyle{p}{{b}}^{{3}}+{\left({q}+{p}\right)}{{b}}^{{2}}+{\left({r}+{q}\right)}{b}+{r}={4}{{b}}^{{3}}+{2}{b}+{6}$

By comparing components,

p = 4

r = 6

q = -4

$\displaystyle{p}{{b}}^{{2}}+{q}{b}+{r}={4}{{b}}^{{2}}-{4}{b}+{6}$

$\displaystyle{4}{{b}}^{{2}}-{4}{b}+{6}={0}$

$\displaystyle{{b}}^{{2}}-{2}{b}+{3}={0}$

delta = $\displaystyle{{\left(-{2}\right)}}^{{2}}-{4}\times{1}\times{3}<{0}$

So this part does not have real solutions. They have complex solutions.

When b<-1 say b = -2 then

dL/db $\displaystyle=\frac{{{4}{{b}}^{{3}}+{2}{b}+{6}}}{{{2}{L}}}=\frac{{-{32}-{4}+{6}}}{{{2}{L}}}$

Since L>0 then dL/db<0.

When b>-1 say b = 0 then

dL/db = $\displaystyle\frac{{{4}{{b}}^{{3}}+{2}{b}+{6}}}{{{2}{L}}}=\frac{{{0}+{0}+{6}}}{{{2}{L}}}$

Since L>0 then dL/db>0.

Since at b= -1 the parabola changes from negative gradient to a positive gradient, that means the curve has a minimum. So the length between points is minimum.

When b = -1 then;

$\displaystyle{a}=-{{b}}^{{2}}$

$\displaystyle{a}=-{1}$

So the closest point to (0,-3) is (-1,-1)which is on x+y^2 = 0 parabola.

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