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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.

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Consider the curve given by the equation \newlinexy2+5xy=50xy^{2}+5xy=50. It can be shown that\newline(dy)/(dx)=(y(y+5))/(x(2y+5))(dy)/(dx)=(-y(y+5))/(x(2y+5)). \newlineWrite the equation of the vertical line that is tangent to the curve.

Full solution

Q. Consider the curve given by the equation \newlinexy2+5xy=50xy^{2}+5xy=50. It can be shown that\newline(dy)/(dx)=(y(y+5))/(x(2y+5))(dy)/(dx)=(-y(y+5))/(x(2y+5)). \newlineWrite the equation of the vertical line that is tangent to the curve.
  1. Set Denominator Equal to Zero: To find the vertical tangent, we need to set the denominator of the derivative equal to 00 because the slope of a vertical line is undefined, which corresponds to an infinite slope or a 00 in the denominator of the derivative.
  2. Solve for xx: Set the denominator of dydx\frac{dy}{dx} equal to zero: x(2y+5)=0x(2y + 5) = 0.
  3. Find x-coordinate: Solve for xx to find the xx-coordinate where the vertical tangent occurs: x=0x = 0.
  4. Equation of Vertical Line: The equation of a vertical line is of the form x=ax = a, where aa is the xx-coordinate of any point on the line.
  5. Equation of Vertical Line: The equation of a vertical line is of the form x=ax = a, where aa is the xx-coordinate of any point on the line.Therefore, the equation of the vertical line that is tangent to the curve is x=0x = 0.

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