Question

\(x^{2}+2 x y+2 y^{2}=45\) \(x+y+x \frac{d y}{d x}+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-(x+y)}{x+2 y}=-2\) \(\Rightarrow x+y=2 x+4 y\) \(\Rightarrow x+3 y=0\) Put eq (II) in eq (I) \(9 y^{2}-6 y^{2}+2 y^{2}=45\) \(y=\pm 3\) \(x=\pm 9\) Pt \(\rightarrow(9,-3), \quad(-9,3)\)

Short answer

Answer: The two points on the curve are (9, -3) and (-9, 3).

Step-by-step solution

Use the given derivative to find an additional equation

We have the expression for the derivative of y with respect to x: \(\frac{d y}{d x}=\frac{-(x+y)}{x+2 y}=-2\) We need to solve this equation to find an equation relating x and y. By cross-multiplication, we get: \(-2(x+2y) = -(x+y)\) This simplifies to: \(x+3y=0\) (Equation II)

Substitute Equation II into Equation I

Now, we have two equations for x and y: Equation I: \(x^{2}+2 x y+2 y^{2}=45\) Equation II: \(x+3y=0\) We can substitute Equation II into Equation I, replacing x with \(-3y\): \((-3y)^2 + 2(-3y)y + 2y^2 = 45\)

Solve for y

Now we solve the quadratic equation we obtained in step 2: \(9y^2 - 6y^2 + 2y^2 = 45\) \(5y^2 = 45\) \(y^2 = 9\) \(y = \pm 3\)

Solve for x

We have found that \(y=\pm 3\). We can now use Equation II (x+3y=0) to solve for x: For \(y=3\): \(x+3(3)=0\) \(x= -9\) For \(y=-3\): \(x+3(-3)=0\) \(x=9\)

Combine results

We have found that: x = -9, y = 3 or x = 9, y = -3 Therefore, the two points on the curve where the derivative is equal to -2 are: (9,-3) and (-9,3).