Consider the curve given by the equation xy^(2)-x^(3)y=8. It can be shown that (dy)/(dx)=(3x^(2)y-y^(2))/(2xy-x^(3)). Write the equation of the vertical line that is tangent to the curve.

Question

Consider the curve given by the equation  xy^(2)-x^(3)y=8. It can be shown that  (dy)/(dx)=(3x^(2)y-y^(2))/(2xy-x^(3)). Write the equation of the vertical line that is tangent to the curve.

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Set Denominator Equal to Zero: To find the vertical tangent, we need to set the denominator of the derivative (dy/dx) equal to zero, because the slope of a vertical line is undefined, and this occurs when the denominator of the slope expression is zero.Factor and Solve for x: The denominator of the derivative is 2xy - x^3. Set this equal to zero to find the x-values where the tangent line could be vertical. 2xy - x^3 = 0Check x = 0: Factor out x from the equation. x(2y - x^2) = 0Solve for y: Setting each factor equal to zero gives us the x-values where the tangent could be vertical. x = 0 or 2y - x^2 = 0Substitute y into Original Equation: If x = 0, then the equation 2y - x^2 = 0 is automatically satisfied, so we don't get additional information from the second factor. We need to check if x = 0 is indeed a point on the curve.Simplify and Combine Terms: Substitute x = 0 into the original equation xy^2 - x^3y = 8 to see if it yields a valid point on the curve. 0*y^2 - 0^3*y = 8 0 = 8 This is not true, so x = 0 is not a point on the curve, and therefore cannot be where a vertical tangent occurs.Solve for x^5: Now, let's solve the second factor 2y - x^2 = 0 for y to find the corresponding y-values for the vertical tangent. 2y = x^2 y = (1/2)x^2No Real Solution: Substitute y = (1/2)x^2 back into the original equation to find the x-values that satisfy both the original equation and the condition for a vertical tangent. x*(1/2*x^2)^2 - x^3*(1/2*x^2) = 8Simplify the equation. x*(1/4*x^4) - (1/2*x^5) = 8 (1/4*x^5) - (1/2*x^5) = 8Combine like terms. -(1/4*x^5) = 8Multiply both sides by -4 to solve for x^5. x^5 = -32Since x^5 = -32 has no real solution (as the fifth root of a negative number is not real), there are no x-values that satisfy the condition for a vertical tangent on the curve xy^2 - x^3y = 8. Therefore, there is no vertical line that is tangent to the curve.

Consider the curve given by the equation $x y^{2}-x^{3} y=8$ . It can be shown that $\frac{d y}{d x}=\frac{3 x^{2} y-y^{2}}{2 x y-x^{3}}$ . $\newline$ Write the equation of the vertical line that is tangent to the curve.

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Consider the curve given by the equation xy2−x3y=8 x y^{2}-x^{3} y=8 xy2−x3y=8. It can be shown that dydx=3x2y−y22xy−x3 \frac{d y}{d x}=\frac{3 x^{2} y-y^{2}}{2 x y-x^{3}} dxdy​=2xy−x33x2y−y2​.\newlineWrite the equation of the vertical line that is tangent to the curve.

Consider the curve given by the equation $x y^{2}-x^{3} y=8$ . It can be shown that $\frac{d y}{d x}=\frac{3 x^{2} y-y^{2}}{2 x y-x^{3}}$ . $\newline$ Write the equation of the vertical line that is tangent to the curve.