Question 1024669
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arc(1/2,-1/3,2sqrt(7/12),-2sqrt(7/36)),
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<pre>This can be done with or without calculus.  I'll do it
both ways since I don't know what math you are taking:

Without using calculus:

Any line through the origin has slope m and y-intercept
(0,0), so its equation is y = mx

If a line is tangent to a curve, the two solutions to
the system of equations of the line and the curve
will be equal.

So we solve the system:

{{{system(x^2+3y^2-x+2y=0,y=mx)}}}

by substitution and set the discriminant = 0,
so that the two solutions will be the same.

{{{x^2+3y^2-x+2y=0}}}
{{{x^2+3(mx)^2-x+2(mx)=0}}}
{{{x^2+3m^2x^2-x+2mx=0}}}
{{{(1+3m^2)x^2+(-1+2m)x=0}}}
{{{(1+3m^2)x^2+(-1+2m)x=0}}}

The discriminant = {{{b^2-4ac}}} = 

{{{(-1+2m)^2-4((1+3m^2)*0)}}} =

{{{(-1+2m)^2}}} =

Set that = 0:

{{{(-1+2m)^2=0}}} =

{{{-1+2m=0}}}

{{{2m=1}}}

{{{m=1/2}}}

So slope = m = {{{1/2}}}

Since the y-intercept is (0,0)

{{{y = mx + b}}}  becomes

{{{y=expr(1/2)x}}}

------------------------

Using calculus:

{{{x^2+3y^2-x+2y=0}}}

Differentiate implicitly:

{{{2x+6y*"y'"-1+2*"y'"=0}}} 

Substitute x=0, y=0

{{{2(0)+6(0)*"y'"-1+2*"y'"=0}}}

{{{-1+2*"y'"=0}}}

{{{2*"y'"=1}}}

{{{"y'"=1/2}}}

So slope = m = {{{1/2}}}

Since the y-intercept is (0,0)

{{{y = mx + b}}}  becomes

{{{y=expr(1/2)x}}}

Edwin</pre>