Question

Find points on the curve \(x^{2}+x y+y^{2}=7\) (a) where the tangent is parallel to the \(x\) -axis and (b) where the tangent is parallel to the \(y\) -axis. (In the latter case, \(d y / d x\) is not defined, but \(d x / d y\) is. What value does \(d x / d y\) have at these points?

Short answer

The points where the tangent is parallel to x-axis are (6, -3) and (1, -1/2). The points where the tangent is parallel to y-axis are (-7/4, 7/2) and (-3/2,3). At these points, the value of \( dx/dy \) is -1.

Step-by-step solution

Find Derivative of Given Function

Differentiate the given equation implicitly with respect to \( x \). The derivative of \( x^{2}+x y+y^{2}=7 \) is \( 2x+y+xy'+2yy' = 0 \). Now, isolate \( y' = dy/dx \) to get \( dy/dx = -(2x+y)/(x+2y) \). The denominator of the expression for \( dy/dx \) gives us the equation for the tangent being parallel to the x-axis (i.e., \( dy/dx = 0 \)), and the numerator gives the equation for the tangent being parallel to the y-axis ( \( dx/dy = 0 \) ).

Find Points for Tangent Parallel to x-axis

For the tangent to the curve to be parallel to the x-axis, \( dy/dx = 0 \). Thus, we have to set the denominator of \( dy/dx \) to zero, i.e., \( x + 2y = 0 \). Solving the given equation and this equation together, we get the points (6, -3) and (1, -1/2) as the points in which the tangent line to the curve parallel to the x-axis.

Find Points for Tangent Parallel to y-axis

For the tangent to the curve to be parallel to the y-axis, \( dx/dy = 0 \). Thus, we have to set the numerator of \( dy/dx \) to zero; i.e., \( 2x+y =0 \). Solving the given equation and this equation together we get the points (-7/4, 7/2) and (-3/2, 3) as the points in which the tangent line to the curve is parallel to the y-axis.

Find the Value of dx/dy

Since \( dx/dy \) is the reciprocal of \( dy/dx \), we have \( dx/dy = -(x+2y)/(2x+y) \). Put the points into this equation, we get \( dx/dy = -1 \) at (-7/4, 7/2) and (-3/2, 3).