# One yes or no and one ellipse equation

• Jun 7th 2006, 05:47 PM
Stuart
One yes or no and one ellipse equation
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?
• Jun 7th 2006, 06:10 PM
ThePerfectHacker
Quote:

Originally Posted by Stuart
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,
$\tau: (x,y)\to (x-4,y-2)$
Thus,
$\frac{x^2}{9}+\frac{y^2}{16}=1$
Becomes,
$\frac{(x-4)^2}{9}+\frac{(y-2)^2}{16}=1$
Thus,
$16(x-4)^2+9(y-2)^2=144$
Thus,
$16x^2-128x+256+9y^2-36y+36=144$
Thus,
$16x^2-128x+9y^2-36y+148=0$
Your mistake it that you used the translation,
$\tau: (x,y)\to (x-4,y+2)$
• Jun 7th 2006, 06:57 PM
Stuart
ok so how would I start the second equation listed?
• Jun 7th 2006, 07:08 PM
ThePerfectHacker
Quote:

Originally Posted by Stuart
ok so how would I start the second equation listed?

By the previous post you have,
$\frac{(x-2-4)^2}{9}+\frac{(y-3-2)^2}{16}=1$
Thus,
$\frac{(x-6)^2}{9}+\frac{(y-5)^2}{16}=1$