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
$$y=(x+1)^2+1$$

Question

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 
$$y=(x+1)^2+1$$

anonymous · Answer

damn typo!

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

mr_perfection_xd · Answer

i wish i was a mathlete

anonymous · Answer

lol me too!

pooja195 · Answer

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

rsadhvika · Answer

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

mr_perfection_xd · Answer

i wanna go beyond the call of duty

geerky42 · Answer

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

geerky42 · Answer

like what I think @rsadhvika is trying to do.

anonymous · Answer

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

anonymous · Answer

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

jim_thompson5910 · Answer

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 $\Large y = (x+1)^2 + 1$

anonymous · Answer

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