Consider the curve given by the equation xy^(2)+5xy=50. It can be shown that (dy)/(dx)=(-y(y+5))/(x(2y+5))". " Write the equation of the vertical line that is tangent to the curve.

Set Denominator Equal to Zero: To find the vertical tangent, we need to set the denominator of the derivative equal to zero because the slope of a vertical line is undefined, which corresponds to an infinite or undefined derivative.Solve for y: Set the denominator of the derivative equal to zero and solve for y: x(2y + 5) = 0 Since x cannot be zero (as it would make the entire original equation 0 = 50, which is false), we must have: 2y + 5 = 0Substitute Back and Solve: Solve for y: 2y = -5 y = -5/2Clear Fraction and Simplify: Now we substitute y = -5/2 back into the original equation to find the corresponding x-values: x(-5/2)^2 + 5x(-5/2) = 50Combine Like Terms: Simplify and solve for x: x(25/4) - (25/2)x = 50 Multiply through by 4 to clear the fraction: 25x - 50x = 200Find x: Combine like terms: -25x = 200Equation of Vertical Line: Solve for x: x = -8The vertical line that is tangent to the curve at the point where y = -5/2 will have an x-coordinate of -8. Therefore, the equation of the vertical line is: x = -8

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Consider the curve given by the equation
xy^(2)+5xy=50. It can be shown that

(dy)/(dx)=(-y(y+5))/(x(2y+5))". "
Write the equation of the vertical line that is tangent to the curve.

Consider the curve given by the equation $x y^{2}+5 x y=50$ . It can be shown that $\frac{d y}{d x}=\frac{-y(y+5)}{x(2 y+5)} \text {. }$ $\newline$ Write the equation of the vertical line that is tangent to the curve.

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Q. Consider the curve given by the equation

x y^{2}+5 x y=50

. It can be shown that

\frac{d y}{d x}=\frac{-y(y+5)}{x(2 y+5)} \text {. }

\newline

Write the equation of the vertical line that is tangent to the curve.

Set Denominator Equal to Zero: To find the vertical tangent, we need to set the denominator of the derivative equal to zero because the slope of a vertical line is undefined, which corresponds to an infinite or undefined derivative.
Solve for y: Set the denominator of the derivative equal to zero and solve for y: $\newline$ $x(2y + 5) = 0$ $\newline$ Since $x$ cannot be zero (as it would make the entire original equation $0 = 50$ , which is false), we must have: $\newline$ $2y + 5 = 0$
Substitute Back and Solve: Solve for $y$ : $2y = -5$ $y = -\frac{5}{2}$
Clear Fraction and Simplify: Now we substitute $y = -\frac{5}{2}$ back into the original equation to find the corresponding $x$ -values: $x\left(-\frac{5}{2}\right)^2 + 5x\left(-\frac{5}{2}\right) = 50$
Combine Like Terms: Simplify and solve for $x$ : $x\left(\frac{25}{4}\right) - \left(\frac{25}{2}\right)x = 50$ Multiply through by $4$ to clear the fraction: $25x - 50x = 200$
Find $x$ : Combine like terms: $-25x = 200$
Equation of Vertical Line: Solve for $x$ : $x = -8$
Equation of Vertical Line: Solve for $x$ : $x = -8$ The vertical line that is tangent to the curve at the point where $y = -\frac{5}{2}$ will have an $x$ -coordinate of $-8$ . Therefore, the equation of the vertical line is: $x = -8$