toxicsugar22 5 years ago 12x^2-3y^2+24y-84=0 give the exact coordinates of any four points on the hyperbola

1. toxicsugar22

hi can u help me

2. dumbcow

if x=0 -3y^2 +24y -84 =0 factor out a -3 y^2 -8y + 28 =0 b^2 -4ac < 0, complex solutions if y=0 12x^2 = 84 x^2 = 7 x = +-sqrt(7) that gives 2 points get x by itself 12x^2 = 3y^2 -24y +84 x^2 = 1/4y^2 - 2y +7 x = sqrt[1/4y^2 -2y +7] pick 2 y_values y=1 --> x =+- sqrt(21/4) = +-sqrt(21)/2 there are other 2 points

3. amistre64

12x^2 -3(y^2 +8y +____) = 84 -3(___) 12x^2 -3(y^2 +8y +16) = 84 -3(16) 12x^2 -3(y-4)^2 = 84 -48 = 36 12/36 x^2 -3/36(y-4)^2 = 1 x^2 (y-4)^2 --- - ------ = 1 3 12

4. toxicsugar22

ok can u help me with one more

5. amistre64

(sqrt(3), 0) (-sqrt(3),0) are 2 that i get

6. amistre64

that might nota worked out ....

7. toxicsugar22

yeah

8. amistre64

i think i got lost trying to type it all in :)

9. toxicsugar22

ok

10. toxicsugar22

but can u help me

11. anonymous

once you have $\frac{x^2}{3}-\frac{(y-4)^2}{12}=1$ cant you just pick y and solve for x?

12. toxicsugar22

y =0

13. toxicsugar22

lets both try that together can u please show me

14. anonymous

say for example $y=2$ $\frac{x^2}{3}-\frac{4}{12}=1$

15. toxicsugar22

ok

16. anonymous

$\frac{x^2}{3}=1+\frac{4}{12}=1+\frac{1}{3}=\frac{4}{3}$

17. dumbcow

i already gave you 4 points

18. anonymous

$x^2=4$ $x=\pm 2$

19. anonymous

or you can pick any other y to solve for x.

20. toxicsugar22

so that is one point

21. anonymous

actually 2: $(-2,4)$ and $(2,4)$

22. dumbcow

(+-sqrt(7), 0) (+-sqrt(21)/2, 1) (+-2, 2) there are 6 points

23. anonymous

let y = 1 $\frac{x^2}{3}-\frac{(1-4)^2}{12}=1$ $\frac{x^2}{3}-\frac{9}{12}=1$ $\frac{x^2}{3}=1+\frac{9}{12}=1+\frac{3}{4}=\frac{7}{4}$

24. anonymous

so $x^2=\frac{21}{4}$ $x=\pm \frac{\sqrt{21}}{2}$

25. anonymous

got it?