# Circle drawn on focal chord of a parabola

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?

• What do you mean by “the” focal chord? There is an infinite number of them. – amd Mar 27 at 19:22
• @amd Any one of the infinite focal chords can be chosen. The product of the parameters will turn out to be same irrespective of which one you take. – Vaishakh Sreekanth Menon Mar 28 at 2:08

It is certainly possible for a circle whose diameter is a focal chord to meet a parabola in four places, although not when that chord is the latus rectum.

The parabola $$4fy=x^2$$ has its focus at $$F=(0,f)$$. A line through $$F$$ with slope $$m$$ meets the parabola at points $$P_{\pm} := f\left(\;2 (m \pm \sqrt{1 + m^2}), 1 + 2 m^2 \pm 2 m \sqrt{1 + m^2}\;\right) \tag{1}$$ The circle with diameter $$\overline{P_{+}P_{-}}$$ has equation $$x^2+y^2- 2 f m x - 2 f y ( 1 + 2 m^2 ) - 3 f^2 = 0 \tag{2}$$ Substituting $$y=x^2/(4f)$$ gives $$\left(x^2-4 f m x -4 f^2\right) \left(x^2+4f m x + 12 f^2 \right) = 0 \tag{3}$$ The roots of the first factor are the $$x$$-coordinates of $$P_{+}$$ and $$P_{-}$$. The roots of the second factor are $$x = 2f \left(-m \pm \sqrt{-3 + m^2}\right) \tag{4}$$ which are real and distinct, leading to two more circle-parabola intersection points, whenever $$|m|>\sqrt{3}$$. (For $$m=\pm \sqrt{3}$$, the two points coincide.) This agrees with @amd's comment to @GReyes' answer. $$\square$$

We observe (either by direct computation from $$(1)$$ and $$(4)$$, or by invoking Vieta's formulas on the constant terms of the factors of $$(3)$$) that the product of the $$x$$-coordinates of $$P_{+}$$ and $$P_{-}$$ is $$-4f^2$$, which is independent of $$m$$; likewise for the product of the $$x$$-coordinates of the other points of intersection, $$12 f^2$$. There's probably a nice geometric interpretation of these facts.

You are right. There are no such real points. If the equation of the parabola is $$y^2=4px$$, the radius of your circle is $$|y(p)|=2p$$. But then, the distance from any point on the circle to the focus is $$2p$$, whereas the distance to the directrix, on the left side of the circle, is less than $$2p$$ (the $$x$$-coordinate of those points is less than $$p$$) and, on the right half, larger than $$2p$$. Therefore, there are no other real intersections with the parabola, since on the parabola the two distances are equal.

• What's described here is only true of the latus rectum. Counterexample: the parabola $y^2=x$, with focus at $(1/4,0)$ and the focal chord drawn through the point $(4,2)$. A circle on this focal chord cuts the parabola in $4$ points. – nickgard Mar 27 at 13:58
• Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-\sqrt3,\sqrt3]$ will generate four intersection points. – amd Mar 27 at 19:40
• I agree. For some reason I understood that the focal chord referred to the latus rectum, probably the use of "the" instead of "a". Not true for a general chord through the focus, of course. – GReyes Mar 27 at 22:19
• But even if we do not chose the latus rectum won't his logic of distances being greater on one side of the circle and less on the other still hold? – Vaishakh Sreekanth Menon Mar 28 at 2:10
• No, it does not work. For one thing, the focus is not the center of the circle anymore. – GReyes Mar 28 at 4:20

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 $$\left(t-\dfrac{1}{t} \right)^2 > 12$$