analytic geometrycirclesconic sections
Is it possible for a circle with diameter as any focal chord of a parabola to cut the parabola at 4 points (2 being the extremities of the focal chord)?
We were asked to find the product of the parameters(t) of the point cut by the circle on the parabola (other than the extremities of the focal chord).Doing some algebra we obtain this product as 3,but I feel that the parameters would be imaginary since I don't think that such a circle can exist.Am I correct?
Let the circle be the unit circle, and take the parabola to have the equation $y-k=a(x-h)^2$, with $a\neq 0$. Note that the slope of the line tangent to the parabola at $(x,y)$ is given by $m = 2 a (x-h)$.
Let $T(\cos\theta,\sin\theta)$ be the point of tangency. (We may assume throughout that $\sin\theta \neq 0$, since our (presumably non-degenerate) parabola has no vertical tangents. Note also that distinct points of intersection must correspond to distinct $x$ coordinates.) Then, as $T$ must satisfy the parabola equation, we have $$\sin\theta - k = a ( \cos\theta - h )^2 \qquad (1)$$ Now, the line tangent to the circle at $T$ necessarily has slope $-\cot\theta$; equating this with the slope relative to the parabola gives $$\cos\theta = -2 a \sin\theta (\cos\theta - h ) \qquad (2)$$
We can solve $(2)$ for $h$, and then $(1)$ for $k$:
$$ h = \frac{\cos\theta \left( 1 + 2 a \sin\theta \right)}{2a\sin\theta} \qquad k = \frac{ 4 a \sin^3\theta - \cos^2\theta }{4a\sin\theta^2}$$ whence the parabola equation becomes
$$y \sin\theta = a x^2 \sin\theta - x \cos\theta \left( 1 + 2 a \sin\theta \right) + 1 + a \cos^2\theta \sin\theta $$
Using that equation to eliminate $y$ in $x^2+y^2=1$ gives this polynomial equation in $x$:
$$\left( x - \cos\theta \right)^2 \left( \left( a \left( x - \cos\theta \right)\sin\theta - \cos\theta\right)^2 + \sin^2\theta \left( 1 + 2 a \sin\theta \right)\right) = 0 \qquad (3) $$
For the second factor of $(3)$ to have any roots requires the quantity $b := 1+2a\sin\theta$ to be non-positive. If $b$ vanishes, we get the double root $x=-\cos\theta$ (the mirror image of $x=\cos\theta$) corresponding to a second point of circle-parabola tangency. For us to have a point of non-tangent intersection, we require $b$ to be strictly negative, and $x=\cos\theta$ to be a root of that second factor. The latter condition implies, by substitution: $$2a\sin^3\theta=-1$$ giving $b=-\cot^2\theta$, which is strictly negative (and defined) for $\theta \neq n \pi/2$. Moreover, equation $(3)$ reduces nicely to $$0 = (x-\cos\theta)^3\left(x-4\cos^3\theta+3\cos\theta\right) = (x-\cos\theta)^3(x-\cos 3\theta)$$ providing the root $x=\cos3\theta$. Double-checking with the parabola equation, we see that the corresponding $y$ value should be $-\sin3\theta$; thus, we can write the second point of intersection as $S(\cos(-3\theta), \sin(-3\theta))$.
The final parabola equation itself is $$ 2 y \sin^3\theta = - \left( x - \cos^3\theta \right)^2 + \sin^4\theta\left( 3 - \sin^2\theta \right)$$
Here's an animation:
(Animation, and interstitial symbol-crunching, courtesy of Mathematica.)
The equation of the circle is given by $$\left(x-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2=\frac 14\left((x_1-x_2)^2+(y_1-y_2)^2\right)$$ From what you've got, you can write it as $$\left(x-\left(a+\frac{2a}{m^2}\right)\right)^2+\left(y-\frac{2a}{m}\right)^2=4a^2\left(\frac{1}{m^2}+1\right)^2$$ Now, set $x=-a$ which is the equation of the directrix to have $$\left(y-\frac{2a}{m}\right)^2=0$$ from which the claim follows.
Best Answer
If $A(at^2,2at)$ is one of the extremities of the focal chord, we know that the other end will be $B\left(\dfrac{a}{t^2}, -\dfrac{2a}{t}\right)$
Let the circle with $AB$ as diameter intersect the parabola at $C(t')$. Since $\angle ACB$ is a right angle, we have $\dfrac{2}{t+t'} \times \dfrac{2}{t'-\dfrac{1}{t}}=-1$
or $t'^2+ \left(t-\dfrac{1}{t} \right)t'+3 =0$
We see that this will have two real roots when $\left(t-\dfrac{1}{t} \right)^2 > 12$