\text{Solve for }x of \left(x-y\right)^2=x+y-1

\begin{align}&x=\frac{2y+1+\sqrt{8y-3}}{2}\\&x=\frac{2y+1-\sqrt{8y-3}}{2}\end{align}

\text{Solve for }y of \left(x-y\right)^2=x+y-1

\begin{align}&y=\frac{2x+1+\sqrt{8x-3}}{2}\\&y=\frac{2x+1-\sqrt{8x-3}}{2}\end{align}

Testing for symmetry about the origin of \left(x-y\right)^2=x+y-1

\textrm{Not symmetry with respect to the origin}

Testing for symmetry about the x-axis of \left(x-y\right)^2=x+y-1

\textrm{Not symmetry with respect to the x-axis}

Testing for symmetry about the y-axis of \left(x-y\right)^2=x+y-1

\textrm{Not symmetry with respect to the y-axis}

Rewrite in polar form of \left(x-y\right)^2=x+y-1

\begin{align}&r=\frac{\cos\left(\theta \right)+\sin\left(\theta \right)+\sqrt{-3+5\sin\left(2\theta \right)}}{2-2\sin\left(2\theta \right)}\\&r=\frac{\cos\left(\theta \right)+\sin\left(\theta \right)-\sqrt{-3+5\sin\left(2\theta \right)}}{2-2\sin\left(2\theta \right)}\end{align}

\text{Find the derivative with respect to }y of \left(x-y\right)^2=x+y-1

\frac{dx}{dy}=\frac{1+2x-2y}{2x-2y-1}

Eliminate xy term by rotating of \left(x-y\right)^2=x+y-1

\left(y^{\prime}\right)^{2}=\frac{\sqrt{2}}{2}\left(x^{\prime}-\frac{1}{\sqrt{2}}\right)

Question :

(x-y)^2=x+y-1

Q: \text{Find the derivative with respect to }x of \left(x-y\right)^2=x+y-1

\frac{dy}{dx}=\frac{1-2x+2y}{-2x+2y-1}

Solve the equation

Solve for x

Solve for y

x = \frac{2 y + 1 + 8 y - 3}{2} x = \frac{2 y + 1 - 8 y - 3}{2}

Show Solution

Testing for symmetry

Testing for symmetry about the origin

Testing for symmetry about the x-axis

Testing for symmetry about the y-axis

Not symmetry with respect to the origin

Show Solution

Rewrite the equation

r = \frac{cos ( θ ) + sin ( θ ) + - 3 + 5 sin ( 2 θ )}{2 - 2 sin ( 2 θ )} r = \frac{cos ( θ ) + sin ( θ ) - - 3 + 5 sin ( 2 θ )}{2 - 2 sin ( 2 θ )}

Show Solution

Find the first derivative

Find the derivative with respect to x

Find the derivative with respect to y

\frac{d y}{d x} = \frac{1 - 2 x + 2 y}{- 2 x + 2 y - 1}

Calculate

(x - y)^{2} = x + y - 1

Take the derivative of both sides

\frac{d}{d x} ((x - y)^{2}) = \frac{d}{d x} (x + y - 1)

Calculate the derivative

More Steps

2 x - 2 y - 2 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = \frac{d}{d x} (x + y - 1)

Calculate the derivative

More Steps

2 x - 2 y - 2 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 1 + \frac{d y}{d x}

Collect like terms by calculating the sum or difference of their coefficients

2 x - 2 y + (- 2 x + 2 y) \frac{d y}{d x} = 1 + \frac{d y}{d x}

Move the expression to the left side

2 x - 2 y + (- 2 x + 2 y) \frac{d y}{d x} - \frac{d y}{d x} = 1

Move the expression to the right side

(- 2 x + 2 y) \frac{d y}{d x} - \frac{d y}{d x} = 1 - (2 x - 2 y)

Collect like terms by calculating the sum or difference of their coefficients

(- 2 x + 2 y - 1) \frac{d y}{d x} = 1 - (2 x - 2 y)

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

(- 2 x + 2 y - 1) \frac{d y}{d x} = 1 - 2 x + 2 y

Divide both sides

\frac{( - 2 x + 2 y - 1 ) \frac{d y}{d x}}{- 2 x + 2 y - 1} = \frac{1 - 2 x + 2 y}{- 2 x + 2 y - 1}

Solution

\frac{d y}{d x} = \frac{1 - 2 x + 2 y}{- 2 x + 2 y - 1}

Show Solution

Conic

(y^{'})^{2} = \frac{2}{2} (x^{'} - \frac{1}{2})

Evaluate

(x - y)^{2} = x + y - 1

Move the expression to the left side

(x - y)^{2} - (x + y - 1) = 0

Calculate

More Steps

x^{2} - 2 x y + y^{2} - x - y + 1 = 0

The coefficients A,B and C of the general equation are A= 1,B= - 2 and C= 1

A = 1 B = - 2 C = 1

To find the angle of rotation θ,substitute the values of A,B and C into the formula cot (2 θ) = \frac{A - C}{B}

cot (2 θ) = \frac{1 - 1}{- 2}

Calculate

cot (2 θ) = 0

Using the unit circle,find the smallest positive angle for which the cotangent is 0

2 θ = \frac{π}{2}

Calculate

θ = \frac{π}{4}

To rotate the axes,use the equation of rotation and substitute \frac{π}{4} for θ

x = x^{'} cos (\frac{π}{4}) - y^{'} sin (\frac{π}{4}) y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} sin (\frac{π}{4}) y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} sin (\frac{π}{4}) + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} \times \frac{2}{2} + y^{'} cos (\frac{π}{4})

Calculate

x = x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2} y = x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}

Substitute x and y into the original equation x^{2} - 2 x y + y^{2} - x - y + 1 = 0

(x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2})^{2} - 2 (x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2}) (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}) + (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2})^{2} - (x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2}) - (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}) + 1 = 0

Calculate

More Steps

Calculate

(x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2})^{2} - 2 (x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2}) (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}) + (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2})^{2} - (x^{'} \times \frac{2}{2} - y^{'} \times \frac{2}{2}) - (x^{'} \times \frac{2}{2} + y^{'} \times \frac{2}{2}) + 1