For ((X^2)/ 9) + ((Y^2)/16) = 1
the general form after 4 units right and 2 units down would be:
16x^2 + 9y^2 - 128x +36y + 148 = 0 Yes/No
and for the equation (((x-2)^2)\4) + (((y-3)^2)/16) = 1
the standard equation would be?
For ((X^2)/ 9) + ((Y^2)/16) = 1
the general form after 4 units right and 2 units down would be:
16x^2 + 9y^2 - 128x +36y + 148 = 0 Yes/No
and for the equation (((x-2)^2)\4) + (((y-3)^2)/16) = 1
the standard equation would be?
The isometry here is a translantion,Originally Posted by Stuart
$\displaystyle \tau: (x,y)\to (x-4,y-2)$
Thus,
$\displaystyle \frac{x^2}{9}+\frac{y^2}{16}=1$
Becomes,
$\displaystyle \frac{(x-4)^2}{9}+\frac{(y-2)^2}{16}=1$
Thus,
$\displaystyle 16(x-4)^2+9(y-2)^2=144$
Thus,
$\displaystyle 16x^2-128x+256+9y^2-36y+36=144$
Thus,
$\displaystyle 16x^2-128x+9y^2-36y+148=0$
Your mistake it that you used the translation,
$\displaystyle \tau: (x,y)\to (x-4,y+2)$