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?

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- Jun 7th 2006, 05:47 PMStuartOne 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 PMThePerfectHackerQuote:

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,

$\displaystyle \tau: (x,y)\to (x-4,y+2)$ - Jun 7th 2006, 06:57 PMStuart
ok so how would I start the second equation listed?

- Jun 7th 2006, 07:08 PMThePerfectHackerQuote:

Originally Posted by**Stuart**

$\displaystyle \frac{(x-2-4)^2}{9}+\frac{(y-3-2)^2}{16}=1$

Thus,

$\displaystyle \frac{(x-6)^2}{9}+\frac{(y-5)^2}{16}=1$