Question

If the origin is shifted to a point \( P \) by the translation of axes to remove the \( y \)-term from the equation \( x^2 - y^2 + 2y - 1 = 0 \), then the transformed equation of it is: 

• A. \( x^2 - y^2 = 1 \)
• B. \( x^2 - y^2 = 0 \)
• C. \( x^2 + y^2 = 1 \)
• D. \( x^2 + y^2 = 0 \)

Hint:

When shifting the origin to eliminate a linear term in a quadratic equation, complete the square and introduce a new coordinate system centered at the new origin.

Answer 1

The Correct Option is B

We are given the equation: \[ x^2 - y^2 + 2y - 1 = 0. \] Step 1: Completing the Square
To eliminate the linear \( y \)-term, complete the square: \[ x^2 - (y^2 - 2y) - 1 = 0. \] Rewriting \( y^2 - 2y \): \[ y^2 - 2y = (y - 1)^2 - 1. \] Thus, substituting back: \[ x^2 - \left( (y - 1)^2 - 1 \right) - 1 = 0. \] \[ x^2 - (y - 1)^2 + 1 - 1 = 0. \] \[ x^2 - (y - 1)^2 = 0. \] Step 2: Shifting the Origin
Introduce a new coordinate system \( Y = y - 1 \), where the origin is shifted to \( (0,1) \). The transformed equation becomes: \[ x^2 - Y^2 = 0. \] Since \( Y = y - 1 \), we conclude: \[ x^2 - y^2 = 0. \] Thus, the transformed equation is: \[ \boxed{x^2 - y^2 = 0.} \]

Answer 2

To eliminate the \( y \)-term from the given equation \( x^2 - y^2 + 2y - 1 = 0 \), we perform a translation of axes. Specifically, we choose a new origin such that the \( y \)-term vanishes.

The given equation is:
\( x^2 - y^2 + 2y - 1 = 0 \)

First, complete the square for the \( y \)-terms:

\( -y^2 + 2y = -(y^2 - 2y) \)

Complete the square inside the parenthesis:
\(-[y^2 - 2y + 1 - 1] = -[(y-1)^2 - 1] = -(y-1)^2 + 1\)

Substitute back into the equation:

\( x^2 - [(y-1)^2 - 1] - 1 = 0 \)

Simplify:

\( x^2 - (y-1)^2 + 1 - 1 = 0 \)
\( x^2 - (y-1)^2 = 0 \)

This implies \( x^2 = (y-1)^2 \), which is already free of the linear \( y \)-term.

The transformed equation of it is:
\( x^2 - y^2 = 0 \)