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Alison tried to find all the points on the curve given by 
x^(2)+4y^(2)=7+3xy where the line tangent to the curve is horizontal. This is her solution:
Step 1: Finding an expression for 
(dy)/(dx).

(dy)/(dx)=(3y-2x)/(8y-3x)
Step 2: Forming a system of equations.

{[x^(2)+4y^(2)=7+3xy],[3y-2x=0],[8y-3x!=0]:}
Step 3: Solving the system.

(3,2) and 
(-3,-2)
Is Alison's solution correct? If not, at which step did she make a mistake?
Choose 1 answer:
(A) The solution is correct.
(B) Step 1 is incorrect.
(C) Step 2 is incorrect.
(D) Step 3 is incorrect.

Alison tried to find all the points on the curve given by x2+4y2=7+3xy x^{2}+4 y^{2}=7+3 x y where the line tangent to the curve is horizontal. This is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=3y2x8y3x \frac{d y}{d x}=\frac{3 y-2 x}{8 y-3 x} \newlineStep 22: Forming a system of equations.\newline{x2+4y2=7+3xy3y2x=08y3x0 \left\{\begin{array}{l} x^{2}+4 y^{2}=7+3 x y \\ 3 y-2 x=0 \\ 8 y-3 x \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newline(3,2) (3,2) and (3,2) (-3,-2) \newlineIs Alison's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.

Full solution

Q. Alison tried to find all the points on the curve given by x2+4y2=7+3xy x^{2}+4 y^{2}=7+3 x y where the line tangent to the curve is horizontal. This is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=3y2x8y3x \frac{d y}{d x}=\frac{3 y-2 x}{8 y-3 x} \newlineStep 22: Forming a system of equations.\newline{x2+4y2=7+3xy3y2x=08y3x0 \left\{\begin{array}{l} x^{2}+4 y^{2}=7+3 x y \\ 3 y-2 x=0 \\ 8 y-3 x \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newline(3,2) (3,2) and (3,2) (-3,-2) \newlineIs Alison's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.
  1. Differentiate implicitly: Find an expression for dydx\frac{dy}{dx}. To find where the tangent line to the curve is horizontal, we need to find where the derivative of yy with respect to xx, dydx\frac{dy}{dx}, is equal to 00. We start by differentiating the given equation implicitly with respect to xx. Differentiating both sides of the equation x2+4y2=7+3xyx^2 + 4y^2 = 7 + 3xy with respect to xx gives us: 2x+8yy=3y+3xy2x + 8yy' = 3y + 3xy'. Rearranging to solve for yy', we get: yy00 yy11 yy22
  2. Form system of equations: Form a system of equations.\newlineFor the tangent to be horizontal, (dydx)(\frac{dy}{dx}) must be equal to 00. This gives us the system of equations:\newline11. x2+4y2=7+3xyx^2 + 4y^2 = 7 + 3xy (original curve equation)\newline22. 3y2x=03y - 2x = 0 (horizontal tangent condition)\newline33. 8y3x08y - 3x \neq 0 (denominator cannot be zero)
  3. Solve the system: Solve the system.\newlineWe need to solve the system of equations from Step 22. Let's start with the second equation:\newline3y2x=03y - 2x = 0\newline3y=2x3y = 2x\newliney=23xy = \frac{2}{3}x\newlineNow we substitute y=23xy = \frac{2}{3}x into the first equation:\newlinex2+4(23x)2=7+3x(23x)x^2 + 4\left(\frac{2}{3}x\right)^2 = 7 + 3x\left(\frac{2}{3}x\right)\newlinex2+169x2=7+2x2x^2 + \frac{16}{9}x^2 = 7 + 2x^2\newlineCombining like terms, we get:\newline259x2=7\frac{25}{9}x^2 = 7\newlinex2=6325x^2 = \frac{63}{25}\newlinex=±6325x = \pm\sqrt{\frac{63}{25}}\newlinex=±375x = \pm\frac{3\sqrt{7}}{5}\newlineNow we find the corresponding y values using y=23xy = \frac{2}{3}x:\newline3y=2x3y = 2x11\newline3y=2x3y = 2x22\newlineThe points where the tangent is horizontal are therefore 3y=2x3y = 2x33 and 3y=2x3y = 2x44.\newlineChecking the third condition, we have:\newline3y=2x3y = 2x55\newline3y=2x3y = 2x66\newline3y=2x3y = 2x77\newline3y=2x3y = 2x88, which is true.

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