Determine the points on the curve 5x² + 6xy +5y² =8 where the tangent is parallel to the line x-y= 1?

Question

Cameron Alexander · Accepted Answer

#(x, y)=(-sqrt2, sqrt2) , (sqrt2, -sqrt2)# #x-y=1=># write in slope-intercept form: #y=x-1=># slope = 1 #5x^2+ 6xy +5y^2 =8=># implicit differentiation: #10x+6y+6xy'+10yy'=0# #y'(3x+5y)=-(5x+3y)# #y'=-(5x+3y)/(3x+5y)=># the slope of the curve, equate to 1: #-(5x+3y)/(3x+5y)=1=># solve for #y# in terms of #x#: #3x+5y=-5x-3y# #8x+8y=0# #x+y=0# #y=-x=># substitute in the equation of the curve, solve for #x#: #5x^2+6x*(-x)+5*(-x)^2=8# #5x^2-6x^2+5x^2=8# #4x^2=8# #x^2=2# #x=-sqrt2 , sqrt2=># solve for #y:# #y=sqrt2 , -sqrt2# hence the required points are: #(x, y)=(-sqrt2, sqrt2) , (sqrt2, -sqrt2)#

Charles Arocha · Answer

undefined To determine the points on the curve $5x^2 + 6xy + 5y^2 = 8$ where the tangent is parallel to the line $x - y = 1$, we need to find the points of tangency.

We start by finding the gradient of the curve and the given line. The gradient of the curve can be found by implicit differentiation. Then, we equate this gradient to the gradient of the given line to find points where the tangents are parallel.

First, we differentiate the equation of the curve implicitly with respect to $x$:

$$ \frac{d}{dx}(5x^2 + 6xy + 5y^2) = \frac{d}{dx}(8) $$

This gives us:

$$ 10x + 6y + 6x\frac{dy}{dx} + 10y\frac{dy}{dx} = 0 $$

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

$$ 6x\frac{dy}{dx} + 10y\frac{dy}{dx} = -10x - 6y $$

$$ \frac{dy}{dx}(6x + 10y) = -10x - 6y $$

$$ \frac{dy}{dx} = \frac{-10x - 6y}{6x + 10y} $$

This is the gradient of the curve.

The gradient of the line $x - y = 1$ is $1$, as it's in the form $y = mx + c$ where $m$ is the slope.

Now, we equate the gradients:

$$ \frac{-10x - 6y}{6x + 10y} = 1 $$

$$ -10x - 6y = 6x + 10y $$

$$ -16x - 16y = 0 $$

$$ x + y = 0 $$

So, the points on the curve where the tangent is parallel to the line $x - y = 1$ are where $x + y = 0$.

To find the corresponding $y$-coordinates, we substitute $x = -y$ into the equation of the curve:

$$ 5(-y)^2 + 6(-y)y + 5y^2 = 8 $$

$$ 5y^2 - 6y^2 + 5y^2 = 8 $$

$$ 4y^2 = 8 $$

$$ y^2 = 2 $$

$$ y = \pm \sqrt{2} $$

So, the points on the curve where the tangent is parallel to the line $x - y = 1$ are $(- \sqrt{2}, \sqrt{2})$ and $(\sqrt{2}, -\sqrt{2})$.