## satellite73 one year ago how do we show, without a picture, that the ellipse $\frac{9+1)^2}{25}+\frac{9y+2)^2}{4}=1$ does not intersect the parabola $y=(x+1)^2+1$

1. anonymous

damn typo!

2. anonymous

$\frac{(x+1)^2}{25}+\frac{(y+2)^2}{4}=1$

3. anonymous

my attempt $y=(x+1)^2+1\\ y-1=(x+1)^2$ so $\frac{y-1}{25}+\frac{(y+2)^2}{4}=1$

4. anonymous

then i multiplied by 100 and get $4(y-1)+25(y+2)^2=100$ but that has a solutions!

5. Mr_Perfection_xD

i wish i was a mathlete

6. anonymous

lol me too!

7. pooja195

.-. well im one and this makes no sense :P

$$y = (x+1)^2+1 \ge 1$$ $$y = -2\pm 2\sqrt{1-\frac{(x+1)^2}{25}} \le -2 \pm 2 = 0$$

9. Mr_Perfection_xD

i wanna go beyond the call of duty

10. geerky42

Maybe somehow check the range of these functions and then we can show that intersection of these ranges is empty set.

11. geerky42

like what I think @rsadhvika is trying to do.

12. anonymous

yeah i graphed and it was obvious but i was looking for algebra i guess the range is the way to go

13. anonymous

probably just should have found the vertices and have been done with it at that point thanks !

14. jim_thompson5910

geerky42 is right on the money solving $$4(y-1)+25(y+2)^2=100$$ gets you these approximate solutions $$\Large y \approx 0.0381$$ $$\Large y \approx -4.1981$$ but neither of these y values are in the range of $$\Large y = (x+1)^2 + 1$$

15. anonymous

yeah minor axis has length 2 and the center is $$(-1,-2)$$ so the whatever they are called are $$(-1,0)$$ and $$(-1,-4)$$