Not sure why this value arises when I am trying to find the tangential intersection?

Question

if we consider the circle, whose equation is given by $$x^2+(y-2)^2=1$$ and the parabola $$y=kx^2$$

We wish to find the values of $k$ for which the parabola will touch the circle (not intersect but touch).

Current Solution

Let us simply substitute for $x^2=\frac{y}{k}$ into the circle, this gives: $$\frac{y}{k}+y^2-4y+4=1 \Rightarrow ky^2+(1-4k)y+3k=0 $$

Now evaluating $$\Delta=0$$ we are solving for the values of $k$ for which the parabola only intersects the circle once.

And so we arrive at $$\Delta=(1-4k)^2-12k=1-8k+16k^2-12k^2=4(k^2-2k)+1$$ $$\Delta=4(k-1)^2-3=0\Rightarrow k=\frac{2\pm\sqrt{3}}{2}$$

However, upon further inspection, the specific value $$k=\frac{2-\sqrt{3}}{2}$$ causes the parabola to never intersect the circle, and thus we only take the other value that we found.

My Problem

While this question is "solved", I do not know why such a value of $k$ rose in the first place? Am I interpreting the discriminant incorrectly?

Because such a value for $k$ will make the parabola touch the circle twice due to of course, the symmetry of this particular example.

Since you substitue that $x^2=y/k$, we need a solution to satisfy $y/k\ge 0$. When $k=(2-\sqrt 3)/2$ this is not true. The superfluous $k$ arises from ignoring the restriction $y/k\ge 0$ in the substitution. — Cave Johnson, Dec 02 '16 at 05:07
But isn't $k=(2-\sqrt{3})/2$ still positive and still satisfies the inequality described? — Hugh Entwistle, Dec 02 '16 at 05:09
$k$ is positve, what about $y$? We need $y/k\ge0$, not $k\ge 0$. — Cave Johnson, Dec 02 '16 at 05:10
And in fact, for $k=(2-\sqrt 3)/2$ the parabola does "touch" the circle, in some special sense: they "touch" at two points where $x$ is a pure imaginary number. — Cave Johnson, Dec 02 '16 at 05:14
I still don't understand how this value of $k$ doesn't satisfy $y/k\geq0$ — Hugh Entwistle, Dec 02 '16 at 05:22
Why does it satify $y/k\ge0$? What's the value of $y$ if $k=(2-\sqrt 3)/2$? — Cave Johnson, Dec 02 '16 at 09:32
Well $y=kx^2$ so the value of $y$ when $k=(2-\sqrt{3})/2$ is $\frac{2-\sqrt{3} }{2x^2} ? $ which is positive? — Hugh Entwistle, Dec 02 '16 at 23:06

score 1 · Answer 1 · answered Dec 04 '16 at 04:01

The step which conceals the introduction of additional, ineligible roots is this one:

Now evaluating $$\Delta=0$$ we are solving for the values of $k$ for which the parabola only intersects the circle once.

The problem here is that $\Delta = 0$ guarantees a root of multiplicity (at least) $2$ to the system of equations, but it does not guarantee that the double root is real. Indeed, the second value of $k$ does in fact correspond to a double root, only that the respective root is complex, outside the real plane, so it doesn't count as an intersection of the real conics.

To follow what happens step by step, take the value $k=\frac{2-\sqrt{3}}{2}$ and replace it back into the equation $ky^2+(1-4k)y+3k=0$. Since it's already established that $\Delta=0$ the double root $y$ is given by:

$$ y = \frac{1-4k}{2k}=\frac{3 - 2 \sqrt{3}}{2 - \sqrt{3}}=(3 - 2 \sqrt{3})(2 + \sqrt{3}) = -\sqrt{3} $$

Then $y= k x^2$ reduces to $x^2 = \frac{y}{k} = \frac{-2\sqrt{3}}{2-\sqrt{3}} = -2\sqrt{3}(2+\sqrt{3}) = -2(3 + 2 \sqrt{3}) \lt 0$ which has no real roots. The complex roots $x=\pm i\,\sqrt{2(3 + 2 \sqrt{3})}$ do in fact correspond to the double roots of the two equations, but they are not "intersections" in the real plane.

score 1 · Answer 2 · answered Dec 04 '16 at 04:09

To find the "lost" solution for $k=(2-\sqrt{3})/2$, plot your curves on a coordinate plane where $x$ is pure imaginary instead of real, while $y$ is still real. What appears as a circle in the all-real plane is now a hyperbola, and the parabola bends down rather than up to become tangent to this hyperbola.

So, the "lost" solution involves imaginary coordinates.

MattG88 · Answer 3 · 2016-12-04T13:05:14.750

0

The problem is that in your substitution, we must require $x^2=\frac{y}{k}\ge0$ and thus:

$${y\over k}=\frac{4k-1\pm\sqrt{4(k^2-2k)+1}}{2k^2}\ge0$$

When $k=\frac{2-\sqrt3}{2}$, $\Delta=0$ so we have one solution but: $$x^2={y\over k}=\frac{4-2\sqrt3-1}{2({2-\sqrt3\over2})^2}<0$$

it is negative hence this solution must be rejected. $\Delta=0$ doesn't ensure $y\ge0$.

edited Dec 04 '16 at 13:05

answered Dec 04 '16 at 03:50

MattG88

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Not sure why this value arises when I am trying to find the tangential intersection?

3 Answers3