# Intersection of Circle and Parabola

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• Nov 17th 2007, 06:22 PM
polymerase
Intersection of Circle and Parabola
A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola $y=2x^2$ tangentially at exactly 2 points. Find these 2 points.

Thanks!
• Nov 17th 2007, 07:47 PM
Jhevon
Quote:

Originally Posted by polymerase
A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola $y=2x^2$ tangentially at exactly 2 points. Find these 2 points.

Thanks!

Let the center of the circle be $(0,a)$ for $a>0$

thus our two curves are:

$x^2 + (y - a)^2 = 1$ and $y = 2x^2$

and we want these two curves to have only two points in common. (it is okay to take this interpretation since the circle will be "inside" the parabola)

plug in $y = 2x^2$ into the equation of the circle, we get:

$x^2 + \left( 2x^2 - a \right)^2 = 1$

$\Rightarrow 4x^4 + (1 - 4a)x^2 + \left( a^2 - 1 \right) = 0$

the above is quadratic in $x^2$

thus, $x^2 = \frac {4a - 1 \pm \sqrt{(1 - 4a)^2 - 16 \left( a^2 - 1 \right) }}8$ ..........................(1)

we want this to have only two solutions, therefore, set the discriminant to zero.

thus, it remains for us to solve:

$(1 - 4a)^2 - 16 \left( a^2 - 1 \right) = 0$ for $a$.

knowing the value for $a$, we plug it into equation (1) to find the x-coordinate for the point of intersection. then use that to get the y-coordinate. (note, one of your solutions for y will be erroneous, be sure to make sure your solutions work in BOTH of the original equations). i leave the rest to you, it's pretty much following your nose from here
• Nov 17th 2007, 08:19 PM
kalagota
Quote:

Originally Posted by polymerase
A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola $y=2x^2$ tangentially at exactly 2 points. Find these 2 points.

Thanks!

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: $x^2 + (y-k)^2 = 1$
$y = 2x^2$

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, $a^2 + (b-k)^2 = 1$ and $b = 2a^2$ and so if we combine both, we have
$a^2 +(2a^2 - k)^2 = 4a^4 - a^2(4k - 1) + k^2 = 1$

or $4a^4 - a^2(4k - 1) + (k^2 - 1) = 0$

using quadratic formula, we have:

$a^2 = \frac{4k-1 \pm \sqrt{(4k - 1)^2 - 16(k^2-1)}}{8} = \frac{4k-1 \pm \sqrt{- 8k + 17}}{8}$

for that to exists, we need $- 8k + 17 \geq 0$ or $k \leq \frac{17}{8}$

in fact, what we need is that $k=\frac{17}{8}$ so that we have a unique solution for $a^2$ and that a has 2 solutions.. Ü so if $k=\frac{17}{8}$, we have $a^2 = \frac{60}{64}$ or $a= \pm \frac{\sqrt{60}}{8}$. and $b=\frac{60}{32}$

hence, the points are $\left( {\frac{\sqrt{60}}{8}, \frac{60}{32}} \right)$ and $\left( {\frac{-\sqrt{60}}{8}, \frac{60}{32}} \right)$

EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe
• Nov 17th 2007, 08:28 PM
Jhevon
Quote:

Originally Posted by kalagota
EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe

hehe, sorry about that. you can do the calculus way if you want. the poster may want to see that since he/she is in calculus
• Nov 18th 2007, 06:38 AM
polymerase
Quote:

Originally Posted by Jhevon
Let the center of the circle be $(0,a)$ for $a>0$

thus our two curves are:

$x^2 + (y - a)^2 = 1$ and $y = 2x^2$

and we want these two curves to have only two points in common. (it is okay to take this interpretation since the circle will be "inside" the parabola)

plug in $y = 2x^2$ into the equation of the circle, we get:

$x^2 + \left( 2x^2 - a \right)^2 = 1$

$\Rightarrow 4x^4 + (1 - 4a)x^2 + \left( a^2 - 1 \right) = 0$

the above is quadratic in $x^2$

thus, $x^2 = \frac {4a - 1 \pm \sqrt{(1 - 4a)^2 - 16 \left( a^2 - 1 \right) }}8$ ..........................(1)

we want this to have only two solutions, therefore, set the discriminant to zero.

thus, it remains for us to solve:

$(1 - 4a)^2 - 16 \left( a^2 - 1 \right) = 0$ for $a$.

knowing the value for $a$, we plug it into equation (1) to find the x-coordinate for the point of intersection. then use that to get the y-coordinate. (note, one of your solutions for y will be erroneous, be sure to make sure your solutions work in BOTH of the original equations). i leave the rest to you, it's pretty much following your nose from here

Thanks for the relay!
• Nov 18th 2007, 06:39 AM
polymerase
Quote:

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: $x^2 + (y-k)^2 = 1$
$y = 2x^2$

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, $a^2 + (b-k)^2 = 1$ and $b = 2a^2$ and so if we combine both, we have
$a^2 +(2a^2 - k)^2 = 4a^4 - a^2(4k - 1) + k^2 = 1$

or $4a^4 - a^2(4k - 1) + (k^2 - 1) = 0$

using quadratic formula, we have:

$a^2 = \frac{4k-1 \pm \sqrt{(4k - 1)^2 - 16(k^2-1)}}{8} = \frac{4k-1 \pm \sqrt{- 8k + 17}}{8}$

for that to exists, we need $- 8k + 17 \geq 0$ or $k \leq \frac{17}{8}$

in fact, what we need is that $k=\frac{17}{8}$ so that we have a unique solution for $a^2$ and that a has 2 solutions.. Ü so if $k=\frac{17}{8}$, we have $a^2 = \frac{60}{64}$ or $a= \pm \frac{\sqrt{60}}{8}$. and $b=\frac{60}{32}$

hence, the points are $\left( {\frac{\sqrt{60}}{8}, \frac{60}{32}} \right)$ and $\left( {\frac{-\sqrt{60}}{8}, \frac{60}{32}} \right)$

EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe

Thanks!