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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 $\newline$ $xy^{2}+5xy=50$ . It can be shown that $\newline$ $(dy)/(dx)=(-y(y+5))/(x(2y+5))$ . $\newline$ Write the equation of the vertical line that is tangent to the curve.

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Find equations of tangent lines using limits

Full solution

Q. Consider the curve given by the equation

\newline

xy^{2}+5xy=50

. It can be shown that

\newline

(dy)/(dx)=(-y(y+5))/(x(2y+5))

.

\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 $0$ because the slope of a vertical line is undefined, which corresponds to an infinite slope or a $0$ in the denominator of the derivative.
Solve for $x$ : Set the denominator of $\frac{dy}{dx}$ equal to zero: $x(2y + 5) = 0$ .
Find x-coordinate: Solve for $x$ to find the $x$ -coordinate where the vertical tangent occurs: $x = 0$ .
Equation of Vertical Line: The equation of a vertical line is of the form $x = a$ , where $a$ is the $x$ -coordinate of any point on the line.
Equation of Vertical Line: The equation of a vertical line is of the form $x = a$ , where $a$ is the $x$ -coordinate of any point on the line.Therefore, the equation of the vertical line that is tangent to the curve is $x = 0$ .

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