Find the point of intersection of the tangents to the curve `y^2 -3xy + x^3 = 3` at the points where `x=-1.`

Wednesday, September 21, 2011

Find the point of intersection of the tangents to the curve $\displaystyle{{y}}^{{2}}-{3}{x}{y}+{{x}}^{{3}}={3}$ at the points where $\displaystyle{x}=-{1}.$

Hello!

The steps are: 1) find the tangent points, 2) find the equations of the tangents, 3) find the intersection of the tangents.

1. It is known that $\displaystyle{x}=-{1}.$ Substitute it to the curve equation and obtain an equation for $\displaystyle{y}:$

$\displaystyle{{y}}^{{2}}+{3}{y}-{1}={3},$ or $\displaystyle{{y}}^{{2}}+{3}{y}-{4}={0}.$

The solutions are $\displaystyle{y}_{{1}}={1}$ and $\displaystyle{y}_{{2}}=-{4}.$ So the points of tangent are $\displaystyle{T}_{{1}}{\left(-{1},{1}\right)}$ and $\displaystyle{T}_{{2}}{\left(-{1},-{4}\right)}.$

2. Differentiate the equation $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}:$

$\displaystyle{2}{y}{y}'-{3}{y}-{3}{x}{y}'+{3}{{x}}^{{2}}={0},$ or $\displaystyle{y}'{\left({2}{y}-{3}{x}\right)}={3}{y}-{3}{{x}}^{{2}},$ or

$\displaystyle{y}'=\frac{{{3}{\left({y}-{{x}}^{{2}}\right)}}}{{{2}{y}-{3}{x}}}.$

For $\displaystyle{T}_{{1}}$ we obtain that $\displaystyle{y}'={0}.$ Thus the tangent is a horizontal line with the equation $\displaystyle{y}={1}.$

For $\displaystyle{T}_{{2}}$ we obtain $\displaystyle{y}'=\frac{{{3}{\left(-{4}-{1}\right)}}}{{{2}\cdot{\left(-{4}\right)}-{3}\cdot{\left(-{1}\right)}}}=-\frac{{15}}{{-{5}}}={3}.$ Thus the equation is $\displaystyle{y}={3}{\left({x}+{1}\right)}-{4}={3}{x}-{1}.$

3. The point of intersection has $\displaystyle{y}={1}$ and $\displaystyle{x}$ from the equation $\displaystyle{1}={3}{x}-{1},$ i.e. $\displaystyle{x}=\frac{{2}}{{3}}.$ So the answer is the point $\displaystyle{\left(\frac{{2}}{{3}},{1}\right)}.$