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