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

Question

Christopher Allers · Accepted Answer

POINTS: #(sqrt(1/3),-2sqrt(1/3)) and (-sqrt(1/3),2sqrt(1/3))# 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'# #2x + (1*y + xy') + 2y*y' = 0 = 2x + y + xy' + 2yy' = 0# STEP 2: We are looking for where #y'=0#, so go ahead and plug 0 in for #y'# in the equation above. #2x + y + x(0) + 2y(0) = 0# #y = -2x# STEP 3: Now we know that we have a horizontal tangent line whenever #y=-2x#. 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=-2x#. STEP 4: When does #y=-2x# on the curve #x^2 + xy + y^2 = 1#? To solve this question, we can sub in #-2x# wherever we see a #y# in our original equation (substitution method). #x^2 + x(-2x) + (-2x)^2 = 1# #x^2 -2x^2 +4x^2 = 1# #3x^2 = 1# #x^2 = 1/3# #x = +- sqrt(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=-2x# where the tangent line is horizontal. POINTS: #(sqrt(1/3),-2sqrt(1/3)) and (-sqrt(1/3),2sqrt(1/3))#

Scarlett Allison · Answer

undefined To find points where the tangent line to the curve $x^2 + xy + y^2 = 1$ is horizontal, we need to find where the derivative of $y$ with respect to $x$ is zero.

First, implicitly differentiate the equation $x^2 + xy + y^2 = 1$ with respect to $x$, then solve for $\frac{dy}{dx}$.

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(1)$

This yields:

$2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx}(x + 2y) = -2x - y$

$\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$

For the tangent line to be horizontal, $\frac{dy}{dx}$ must be equal to zero. So, we have:

$\frac{-2x - y}{x + 2y} = 0$

This implies that $-2x - y = 0$, which simplifies to $y = -2x$.

Now, substitute $y = -2x$ into the original equation $x^2 + xy + y^2 = 1$:

$x^2 + x(-2x) + (-2x)^2 = 1$

$x^2 - 2x^2 + 4x^2 = 1$

$3x^2 = 1$

$x^2 = \frac{1}{3}$

$x = \pm \sqrt{\frac{1}{3}}$

Now, plug these values of $x$ into $y = -2x$ to find the corresponding $y$ values:

For $x = \sqrt{\frac{1}{3}}$, $y = -2\sqrt{\frac{1}{3}}$

For $x = -\sqrt{\frac{1}{3}}$, $y = 2\sqrt{\frac{1}{3}}$

Therefore, the points where the tangent line to the curve $x^2 + xy + y^2 = 1$ is horizontal are $\left(\sqrt{\frac{1}{3}}, -2\sqrt{\frac{1}{3}}\right)$ and $\left(-\sqrt{\frac{1}{3}}, 2\sqrt{\frac{1}{3}}\right)$.