Circle drawn on focal chord of a parabolaCircle Chord SequenceIntersection of parabola and circleLongest chord inside the intersection area of three circlesMaximum product of lengths involving secant drawn to a parabola.Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as diameter.Show that the circle drawn on a focal chord of a parabola $y^2=4ax$, as a diameter touches the directrixUnable to prove a statement regarding circles and trigonometry.Converse of a theorem for parabolasChord tangent to two circlesPossibly wrong question in S L Loney Coordinate Geometry
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Circle drawn on focal chord of a parabola
Circle Chord SequenceIntersection of parabola and circleLongest chord inside the intersection area of three circlesMaximum product of lengths involving secant drawn to a parabola.Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as diameter.Show that the circle drawn on a focal chord of a parabola $y^2=4ax$, as a diameter touches the directrixUnable to prove a statement regarding circles and trigonometry.Converse of a theorem for parabolasChord tangent to two circlesPossibly wrong question in S L Loney Coordinate Geometry
$begingroup$
Is it possible for a circle with diameter as the 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?
analytic-geometry circles conic-sections
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
$endgroup$
add a comment |
$begingroup$
Is it possible for a circle with diameter as the 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?
analytic-geometry circles conic-sections
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
$endgroup$
$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago
add a comment |
$begingroup$
Is it possible for a circle with diameter as the 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?
analytic-geometry circles conic-sections
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
$endgroup$
Is it possible for a circle with diameter as the 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?
analytic-geometry circles conic-sections
analytic-geometry circles conic-sections
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
asked 15 hours ago
Vaishakh Sreekanth MenonVaishakh Sreekanth Menon
293
293
$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago
add a comment |
$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago
$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago
$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 15 hours ago
GReyesGReyes
2,31315
$endgroup$
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
add a comment |
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1 Answer
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$begingroup$
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.
answered 15 hours ago
GReyesGReyes
2,31315
$endgroup$
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
add a comment |
$begingroup$
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.
answered 15 hours ago
GReyesGReyes
2,31315
$endgroup$
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
add a comment |
$begingroup$
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.
answered 15 hours ago
GReyesGReyes
2,31315
$endgroup$
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.
answered 15 hours ago
GReyesGReyes
2,31315
answered 15 hours ago
GReyesGReyes
2,31315
answered 15 hours ago
GReyesGReyes
2,31315
answered 15 hours ago
GReyesGReyes
2,31315
2,31315
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
add a comment |
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
1
1
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
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.
$endgroup$
– nickgard
8 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-sqrt3,sqrt3]$ will generate four intersection points.
$endgroup$
– amd
2 hours ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
$begingroup$
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.
$endgroup$
– GReyes
16 mins ago
add a comment |
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$begingroup$
What do you mean by “the” focal chord? There is an infinite number of them.
$endgroup$
– amd
3 hours ago