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A double ordinate of the parabola
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A double ordinate of the parabola $y^2=2ax $ is of length $4a $. Prove that the lines joining the vertex to its ends are at right angles.
What does double ordinate actually mean? Seeing that the length is $4a $, I guess it is Latus Rectum. How do I proceed this?
analytic-geometry conic-sections
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
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A double ordinate of the parabola $y^2=2ax $ is of length $4a $. Prove that the lines joining the vertex to its ends are at right angles.
What does double ordinate actually mean? Seeing that the length is $4a $, I guess it is Latus Rectum. How do I proceed this?
analytic-geometry conic-sections
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36
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up vote
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up vote
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favorite
A double ordinate of the parabola $y^2=2ax $ is of length $4a $. Prove that the lines joining the vertex to its ends are at right angles.
What does double ordinate actually mean? Seeing that the length is $4a $, I guess it is Latus Rectum. How do I proceed this?
analytic-geometry conic-sections
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
A double ordinate of the parabola $y^2=2ax $ is of length $4a $. Prove that the lines joining the vertex to its ends are at right angles.
What does double ordinate actually mean? Seeing that the length is $4a $, I guess it is Latus Rectum. How do I proceed this?
analytic-geometry conic-sections
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
asked Jul 25 at 13:33
blue_eyed_...
3,19321136
3,19321136
I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36
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I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36
I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36
I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36
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3 Answers
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So if $y= pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
answered Jul 25 at 13:40
greedoid
26.1k93473
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
add a comment |Â
up vote
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If $x=2at^2, y^2=(2at)^2implies y=pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$dfrac2a(q-p)2a(q^2-p^2)=dfrac1q+p$$
But it's perpendicular to the axis $y=0$ of the parabola
$implies q+p=0iff q=-p$
Now the length of $PQ$ will be $$sqrt(2ap^2-2aq^2)^2+(2ap-2aq)^2=4a|q|$$
Finally, if $O(0,0)$ is vertex,
what will be product of gradients of $OP,OQ$?
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
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Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)mbox and P^prime=(at^2,-2at)$$
$$PP^prime=4a$$
$$sqrt(at^2-at^2)^2+(2at+2at)^2=4a$$
$$sqrt(4at)^2=4a$$
$$4at=4aimplies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^prime=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^prime=m_2=-dfrac 12$
$$implies m_1times m_2=-1$$
Hence $angle POP^prime=dfracpi2$
Therefore, they form right triangles.
answered Jul 25 at 15:08
Key Flex
4,193423
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
So if $y= pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
answered Jul 25 at 13:40
greedoid
26.1k93473
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
add a comment |Â
up vote
2
down vote
So if $y= pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
answered Jul 25 at 13:40
greedoid
26.1k93473
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
So if $y= pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
answered Jul 25 at 13:40
greedoid
26.1k93473
So if $y= pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
answered Jul 25 at 13:40
greedoid
26.1k93473
answered Jul 25 at 13:40
greedoid
26.1k93473
answered Jul 25 at 13:40
greedoid
26.1k93473
answered Jul 25 at 13:40
greedoid
26.1k93473
26.1k93473
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
add a comment |Â
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
what does double ordinate mean?
– blue_eyed_...
Jul 25 at 13:41
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
It is ordinate times 2.
– greedoid
Jul 25 at 13:42
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
blogformathematics.blogspot.com/2014/04/…
– greedoid
Jul 25 at 13:44
add a comment |Â
up vote
1
down vote
If $x=2at^2, y^2=(2at)^2implies y=pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$dfrac2a(q-p)2a(q^2-p^2)=dfrac1q+p$$
But it's perpendicular to the axis $y=0$ of the parabola
$implies q+p=0iff q=-p$
Now the length of $PQ$ will be $$sqrt(2ap^2-2aq^2)^2+(2ap-2aq)^2=4a|q|$$
Finally, if $O(0,0)$ is vertex,
what will be product of gradients of $OP,OQ$?
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
add a comment |Â
up vote
1
down vote
If $x=2at^2, y^2=(2at)^2implies y=pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$dfrac2a(q-p)2a(q^2-p^2)=dfrac1q+p$$
But it's perpendicular to the axis $y=0$ of the parabola
$implies q+p=0iff q=-p$
Now the length of $PQ$ will be $$sqrt(2ap^2-2aq^2)^2+(2ap-2aq)^2=4a|q|$$
Finally, if $O(0,0)$ is vertex,
what will be product of gradients of $OP,OQ$?
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $x=2at^2, y^2=(2at)^2implies y=pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$dfrac2a(q-p)2a(q^2-p^2)=dfrac1q+p$$
But it's perpendicular to the axis $y=0$ of the parabola
$implies q+p=0iff q=-p$
Now the length of $PQ$ will be $$sqrt(2ap^2-2aq^2)^2+(2ap-2aq)^2=4a|q|$$
Finally, if $O(0,0)$ is vertex,
what will be product of gradients of $OP,OQ$?
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
If $x=2at^2, y^2=(2at)^2implies y=pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$dfrac2a(q-p)2a(q^2-p^2)=dfrac1q+p$$
But it's perpendicular to the axis $y=0$ of the parabola
$implies q+p=0iff q=-p$
Now the length of $PQ$ will be $$sqrt(2ap^2-2aq^2)^2+(2ap-2aq)^2=4a|q|$$
Finally, if $O(0,0)$ is vertex,
what will be product of gradients of $OP,OQ$?
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
answered Jul 25 at 13:59
lab bhattacharjee
215k14152264
215k14152264
add a comment |Â
add a comment |Â
up vote
1
down vote
Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)mbox and P^prime=(at^2,-2at)$$
$$PP^prime=4a$$
$$sqrt(at^2-at^2)^2+(2at+2at)^2=4a$$
$$sqrt(4at)^2=4a$$
$$4at=4aimplies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^prime=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^prime=m_2=-dfrac 12$
$$implies m_1times m_2=-1$$
Hence $angle POP^prime=dfracpi2$
Therefore, they form right triangles.
answered Jul 25 at 15:08
Key Flex
4,193423
add a comment |Â
up vote
1
down vote
Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)mbox and P^prime=(at^2,-2at)$$
$$PP^prime=4a$$
$$sqrt(at^2-at^2)^2+(2at+2at)^2=4a$$
$$sqrt(4at)^2=4a$$
$$4at=4aimplies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^prime=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^prime=m_2=-dfrac 12$
$$implies m_1times m_2=-1$$
Hence $angle POP^prime=dfracpi2$
Therefore, they form right triangles.
answered Jul 25 at 15:08
Key Flex
4,193423
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)mbox and P^prime=(at^2,-2at)$$
$$PP^prime=4a$$
$$sqrt(at^2-at^2)^2+(2at+2at)^2=4a$$
$$sqrt(4at)^2=4a$$
$$4at=4aimplies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^prime=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^prime=m_2=-dfrac 12$
$$implies m_1times m_2=-1$$
Hence $angle POP^prime=dfracpi2$
Therefore, they form right triangles.
answered Jul 25 at 15:08
Key Flex
4,193423
Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)mbox and P^prime=(at^2,-2at)$$
$$PP^prime=4a$$
$$sqrt(at^2-at^2)^2+(2at+2at)^2=4a$$
$$sqrt(4at)^2=4a$$
$$4at=4aimplies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^prime=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^prime=m_2=-dfrac 12$
$$implies m_1times m_2=-1$$
Hence $angle POP^prime=dfracpi2$
Therefore, they form right triangles.
answered Jul 25 at 15:08
Key Flex
4,193423
answered Jul 25 at 15:08
Key Flex
4,193423
answered Jul 25 at 15:08
Key Flex
4,193423
answered Jul 25 at 15:08
Key Flex
4,193423
4,193423
add a comment |Â
add a comment |Â
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I guess if $x=2a$ then $y^2 =4a^2$ so $y=pm 2a$, so $|y_1-y_2| =4a$
– greedoid
Jul 25 at 13:36