Possibly wrong question in S L Loney Coordinate Geometry
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Given question:
$P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$
What I did:
I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome
$$RM = a(t_1^2-t_3^2)$$
$$RN = a(t_2^2-t_3^2)$$
$$RV = -a(t_1-t_3)(t_2-t_3)$$
Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong?
analytic-geometry
asked Nov 12 at 9:04
Shubhraneel Pal
34029
add a comment |
up vote
0
down vote
favorite
Given question:
$P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$
What I did:
I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome
$$RM = a(t_1^2-t_3^2)$$
$$RN = a(t_2^2-t_3^2)$$
$$RV = -a(t_1-t_3)(t_2-t_3)$$
Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong?
analytic-geometry
asked Nov 12 at 9:04
Shubhraneel Pal
34029
Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given question:
$P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$
What I did:
I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome
$$RM = a(t_1^2-t_3^2)$$
$$RN = a(t_2^2-t_3^2)$$
$$RV = -a(t_1-t_3)(t_2-t_3)$$
Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong?
analytic-geometry
asked Nov 12 at 9:04
Shubhraneel Pal
34029
Given question:
$P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$
What I did:
I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome
$$RM = a(t_1^2-t_3^2)$$
$$RN = a(t_2^2-t_3^2)$$
$$RV = -a(t_1-t_3)(t_2-t_3)$$
Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong?
analytic-geometry
analytic-geometry
asked Nov 12 at 9:04
Shubhraneel Pal
34029
asked Nov 12 at 9:04
Shubhraneel Pal
34029
asked Nov 12 at 9:04
Shubhraneel Pal
34029
asked Nov 12 at 9:04
Shubhraneel Pal
34029
asked Nov 12 at 9:04
Shubhraneel Pal
34029
34029
Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56
add a comment |
Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56
Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
In a parabola the abscissa is proportional to the square or the related ordinate, that is:
$$
RM=kPM^2,quad RN=kQN^2.
$$
On the other hand, by similar triangles we have:
$$
begin{align}
VM/VN &= PM/QN \
(VM+VN)/VN &= (PM+QN)/QN \
(RN-RM)/VN &= (PM+QN)/QN \
k(QN^2-PM^2)/VN &= (PM+QN)/QN \
k(QN-PM)QN &= VN \
kQN^2-kPMcdot QN &= VN \
RN-kPMcdot QN &= VN \
RN-VN &= kPMcdot QN \
RV &= kPMcdot QN.
end{align}
$$
Hence:
$$
RV^2 = kPM^2cdot kQN^2 =RMcdot RN.
$$
answered Nov 12 at 20:43
Aretino
22.3k21442
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In a parabola the abscissa is proportional to the square or the related ordinate, that is:
$$
RM=kPM^2,quad RN=kQN^2.
$$
On the other hand, by similar triangles we have:
$$
begin{align}
VM/VN &= PM/QN \
(VM+VN)/VN &= (PM+QN)/QN \
(RN-RM)/VN &= (PM+QN)/QN \
k(QN^2-PM^2)/VN &= (PM+QN)/QN \
k(QN-PM)QN &= VN \
kQN^2-kPMcdot QN &= VN \
RN-kPMcdot QN &= VN \
RN-VN &= kPMcdot QN \
RV &= kPMcdot QN.
end{align}
$$
Hence:
$$
RV^2 = kPM^2cdot kQN^2 =RMcdot RN.
$$
answered Nov 12 at 20:43
Aretino
22.3k21442
add a comment |
up vote
3
down vote
accepted
In a parabola the abscissa is proportional to the square or the related ordinate, that is:
$$
RM=kPM^2,quad RN=kQN^2.
$$
On the other hand, by similar triangles we have:
$$
begin{align}
VM/VN &= PM/QN \
(VM+VN)/VN &= (PM+QN)/QN \
(RN-RM)/VN &= (PM+QN)/QN \
k(QN^2-PM^2)/VN &= (PM+QN)/QN \
k(QN-PM)QN &= VN \
kQN^2-kPMcdot QN &= VN \
RN-kPMcdot QN &= VN \
RN-VN &= kPMcdot QN \
RV &= kPMcdot QN.
end{align}
$$
Hence:
$$
RV^2 = kPM^2cdot kQN^2 =RMcdot RN.
$$
answered Nov 12 at 20:43
Aretino
22.3k21442
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In a parabola the abscissa is proportional to the square or the related ordinate, that is:
$$
RM=kPM^2,quad RN=kQN^2.
$$
On the other hand, by similar triangles we have:
$$
begin{align}
VM/VN &= PM/QN \
(VM+VN)/VN &= (PM+QN)/QN \
(RN-RM)/VN &= (PM+QN)/QN \
k(QN^2-PM^2)/VN &= (PM+QN)/QN \
k(QN-PM)QN &= VN \
kQN^2-kPMcdot QN &= VN \
RN-kPMcdot QN &= VN \
RN-VN &= kPMcdot QN \
RV &= kPMcdot QN.
end{align}
$$
Hence:
$$
RV^2 = kPM^2cdot kQN^2 =RMcdot RN.
$$
answered Nov 12 at 20:43
Aretino
22.3k21442
In a parabola the abscissa is proportional to the square or the related ordinate, that is:
$$
RM=kPM^2,quad RN=kQN^2.
$$
On the other hand, by similar triangles we have:
$$
begin{align}
VM/VN &= PM/QN \
(VM+VN)/VN &= (PM+QN)/QN \
(RN-RM)/VN &= (PM+QN)/QN \
k(QN^2-PM^2)/VN &= (PM+QN)/QN \
k(QN-PM)QN &= VN \
kQN^2-kPMcdot QN &= VN \
RN-kPMcdot QN &= VN \
RN-VN &= kPMcdot QN \
RV &= kPMcdot QN.
end{align}
$$
Hence:
$$
RV^2 = kPM^2cdot kQN^2 =RMcdot RN.
$$
answered Nov 12 at 20:43
Aretino
22.3k21442
edited Nov 13 at 7:08
answered Nov 12 at 20:43
Aretino
22.3k21442
answered Nov 12 at 20:43
Aretino
22.3k21442
answered Nov 12 at 20:43
Aretino
22.3k21442
22.3k21442
add a comment |
add a comment |
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Can you draw a clear picture of the given problem on paper and post a snapshot although I know it yields to downvoting and some frowning people commenting you not to do so.. This sometimes helps us solve.
– Saradamani
Nov 12 at 9:23
"Ordinates $PM$ and $QN$ are drawn to this diameter". I think you misinterpreted this: an ordinate to a diameter is a line, parallel to the tangent at the intersection between that diameter and the ellipse.
– Aretino
Nov 12 at 19:09
@Aretino yeah I was suspecting that I possibly haven't taken the meaning of the question right
– Shubhraneel Pal
Nov 13 at 8:56