Provide an appropriate response.
-At the two points where the curve $x ^ { 2 } - x y + y ^ { 2 } = 4$ crosses the $x$-axis, the tangents to the curve are parallel. What is the common slope of these tangents?
A) 1
B) -1
C) 2
D) -2

Question

Quizplus · Accepted Answer

To find the slope of the tangent lines at the points where the curve crosses the x-axis, we need to take the derivative of the equation with respect to x:

$$\frac{d}{dx}(x^2-xy+y^2)= \frac{d}{dx}4$$
$$2x -y -x\frac{dy}{dx}+ 2y\frac{dy}{dx} = 0$$
$$(2y-x\frac{dy}{dx}) = (2x-y) \frac{dy}{dx}$$
$$\frac{dy}{dx} = \frac{2y-x\frac{dy}{dx}}{2x-y}$$

At the points where the curve crosses the x-axis, y=0, so we have:

$$\frac{dy}{dx} = \frac{-x \frac{dy}{dx}}{2x}$$
$$1 = \frac{-\frac{dy}{dx}}{2}$$
$$\frac{dy}{dx} = -2$$

Therefore, the common slope of the tangent lines at the two points where the curve crosses the x-axis is -2.

Provide an Appropriate Response x2−xy+y2=4x ^ { 2 } - x y + y ^ { 2 } = 4x2−xy+y2=4

Provide an Appropriate Response $x ^ { 2 } - x y + y ^ { 2 } = 4$