After translation of coordinate axes, a quadratic
equation becomes $\frac{\left(x^{\prime}\right)^{2}}{16}+\frac{\left(y^{\prime}\right)^{2}}{9}=1$
had the effect of moving the point (x, y)=(1,-2) to
the point $\left(x^{\prime}, y^{\prime}\right)=(0,0)$
equation in x y coordinates?
A) $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
B) $\frac{(x+1)^{2}}{16}+\frac{(y-2)^{2}}{9}=1$
C) $\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1$
D) $\frac{\left(x^{\prime}-1\right)^{2}}{16}+\frac{\left(y^{\prime}+2\right)^{2}}{9}=1$

Question

After translation of coordinate axes, a quadratic
equation becomes $\frac{\left(x^{\prime}\right)^{2}}{16}+\frac{\left(y^{\prime}\right)^{2}}{9}=1$
had the effect of moving the point  (x, y)=(1,-2)  to
the point $\left(x^{\prime}, y^{\prime}\right)=(0,0)$
equation in  x y  coordinates?
A) $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
B) $\frac{(x+1)^{2}}{16}+\frac{(y-2)^{2}}{9}=1$
C) $\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1$
D) $\frac{\left(x^{\prime}-1\right)^{2}}{16}+\frac{\left(y^{\prime}+2\right)^{2}}{9}=1$

Quizplus · Accepted Answer

The translation of the coordinate axes that moves the point (1, -2) to (0, 0) implies that the original coordinates (x, y) are related to the new coordinates (x', y') by the equations x' = x - 1 and y' = y + 2. Substituting these into the given equation in x'y' coordinates gives the equation in xy coordinates as $\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1$.

After Translation of Coordinate Axes, a Quadratic Equation Becomes (x′)216+(y′)29=1\frac{\left(x^{\prime}\right)^{2}}{16}+\frac{\left(y^{\prime}\right)^{2}}{9}=116(x′)2​+9(y′)2​=1 Had the Effect of Moving the Point (X, Y)=(1,-2)

After Translation of Coordinate Axes, a Quadratic
Equation Becomes $\frac{\left(x^{\prime}\right)^{2}}{16}+\frac{\left(y^{\prime}\right)^{2}}{9}=1$ Had the Effect of Moving the Point (X, Y)=(1,-2)