Originally Posted by
kalagota i have solved this for my tutor before but i can't find my solutions..
but i think, this can be solved in two ways, one is using calculus while the other is using some college algebra..
let (0,k), k>0 be the center of the circle: $\displaystyle x^2 + (y-k)^2 = 1$
$\displaystyle y = 2x^2$
let us use the fact that both are symmetric wrt to y axis.. hence, let (a,b) and (-a,b) be the points of intersection..
so, $\displaystyle a^2 + (b-k)^2 = 1$ and $\displaystyle b = 2a^2$ and so if we combine both, we have
$\displaystyle a^2 +(2a^2 - k)^2 = 4a^4 - a^2(4k - 1) + k^2 = 1$
or $\displaystyle 4a^4 - a^2(4k - 1) + (k^2 - 1) = 0$
using quadratic formula, we have:
$\displaystyle a^2 = \frac{4k-1 \pm \sqrt{(4k - 1)^2 - 16(k^2-1)}}{8} = \frac{4k-1 \pm \sqrt{- 8k + 17}}{8}$
for that to exists, we need $\displaystyle - 8k + 17 \geq 0$ or $\displaystyle k \leq \frac{17}{8}$
in fact, what we need is that $\displaystyle k=\frac{17}{8}$ so that we have a unique solution for $\displaystyle a^2$ and that a has 2 solutions.. Ü so if $\displaystyle k=\frac{17}{8}$, we have $\displaystyle a^2 = \frac{60}{64}$ or $\displaystyle a= \pm \frac{\sqrt{60}}{8}$. and $\displaystyle b=\frac{60}{32}$
hence, the points are $\displaystyle \left( {\frac{\sqrt{60}}{8}, \frac{60}{32}} \right)$ and $\displaystyle \left( {\frac{-\sqrt{60}}{8}, \frac{60}{32}} \right)$
EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe