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:
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:
\( x^2 - y^2 = 0 \)
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |