# Find all points where the tangent line is horizontal: x^2+xy+y^2=1?

Oct 23, 2016

POINTS: $\left(\sqrt{\frac{1}{3}} , - 2 \sqrt{\frac{1}{3}}\right) \mathmr{and} \left(- \sqrt{\frac{1}{3}} , 2 \sqrt{\frac{1}{3}}\right)$

#### Explanation:

We know the tangent line is horizontal when $y ' = 0$. So we want to find all points on the curve where $y ' = 0$.

STEP 1: Use implicit differentiation to find $y '$
$2 x + \left(1 \cdot y + x y '\right) + 2 y \cdot y ' = 0 = 2 x + y + x y ' + 2 y y ' = 0$

STEP 2: We are looking for where $y ' = 0$, so go ahead and plug 0 in for $y '$ in the equation above.
$2 x + y + x \left(0\right) + 2 y \left(0\right) = 0$
$y = - 2 x$

STEP 3: Now we know that we have a horizontal tangent line whenever $y = - 2 x$. But the question is asking us "for what points." To find the points, we are looking for the points on the curve for which $y = - 2 x$.

STEP 4: When does $y = - 2 x$ on the curve ${x}^{2} + x y + {y}^{2} = 1$?
To solve this question, we can sub in $- 2 x$ wherever we see a $y$ in our original equation (substitution method).
${x}^{2} + x \left(- 2 x\right) + {\left(- 2 x\right)}^{2} = 1$
${x}^{2} - 2 {x}^{2} + 4 {x}^{2} = 1$
$3 {x}^{2} = 1$
${x}^{2} = \frac{1}{3}$
$x = \pm \sqrt{\frac{1}{3}}$

STEP 5: Now that we know the x-value of the point, we can easily find the y-value of the point because we know $y = - 2 x$ where the tangent line is horizontal.

POINTS: $\left(\sqrt{\frac{1}{3}} , - 2 \sqrt{\frac{1}{3}}\right) \mathmr{and} \left(- \sqrt{\frac{1}{3}} , 2 \sqrt{\frac{1}{3}}\right)$