Coordinate Geometry: Locus
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Problem: With a given point and a line as focus and directrix, a series of ellipses are described, what is the locus of the extremities of their minor axes?
I tried this problem with pure geometry but could not solve. Please help me in this.
geometry analytic-geometry conic-sections
$endgroup$
add a comment |
$begingroup$
Problem: With a given point and a line as focus and directrix, a series of ellipses are described, what is the locus of the extremities of their minor axes?
I tried this problem with pure geometry but could not solve. Please help me in this.
geometry analytic-geometry conic-sections
$endgroup$
$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11
add a comment |
$begingroup$
Problem: With a given point and a line as focus and directrix, a series of ellipses are described, what is the locus of the extremities of their minor axes?
I tried this problem with pure geometry but could not solve. Please help me in this.
geometry analytic-geometry conic-sections
$endgroup$
Problem: With a given point and a line as focus and directrix, a series of ellipses are described, what is the locus of the extremities of their minor axes?
I tried this problem with pure geometry but could not solve. Please help me in this.
geometry analytic-geometry conic-sections
geometry analytic-geometry conic-sections
edited Dec 20 '18 at 13:26
jayant98
625216
625216
asked Dec 20 '18 at 9:49
prashant sharmaprashant sharma
757
757
$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11
add a comment |
$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11
$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using polar coordinates:
$$r=frac{ep}{1-ecos theta}=frac{b^2}{a-ccos theta}$$
The extremities of the minor axis are:
$$(x,y)=(c,pm b)$$
Since
$$p=frac{b^2}{c}$$
The required locus is
$$fbox{$y^2=px$}$$
$endgroup$
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
|
show 1 more comment
$begingroup$
I am trying a shortcut but stuck up..
Property of ellipse $ (x/a)^2+(y/b)^2 =1$ as sum of focal chord lengths is $2a$
$$ sqrt{x^2+y^2} + sqrt{(x-2c)^2+y^2} = 2 a = 2 sqrt{b^2+c^2} $$
Shifting origin to ellipse focus we can put
$$ c=x;, b=y $$
plug in and simplify the mix-up situation
$$ x^2-4cx+y^2 +y^2 = x^2+y^2 quadrightarrow y^2= 2 c x $$
If $p$ is directrix-focus distance Ng Chung Tak gets
$$ y^2= p x $$
I am immediately unable to figure out under what legitimate/interpretative conditions $p= 2c$ holds good,
$endgroup$
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using polar coordinates:
$$r=frac{ep}{1-ecos theta}=frac{b^2}{a-ccos theta}$$
The extremities of the minor axis are:
$$(x,y)=(c,pm b)$$
Since
$$p=frac{b^2}{c}$$
The required locus is
$$fbox{$y^2=px$}$$
$endgroup$
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
|
show 1 more comment
$begingroup$
Using polar coordinates:
$$r=frac{ep}{1-ecos theta}=frac{b^2}{a-ccos theta}$$
The extremities of the minor axis are:
$$(x,y)=(c,pm b)$$
Since
$$p=frac{b^2}{c}$$
The required locus is
$$fbox{$y^2=px$}$$
$endgroup$
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
|
show 1 more comment
$begingroup$
Using polar coordinates:
$$r=frac{ep}{1-ecos theta}=frac{b^2}{a-ccos theta}$$
The extremities of the minor axis are:
$$(x,y)=(c,pm b)$$
Since
$$p=frac{b^2}{c}$$
The required locus is
$$fbox{$y^2=px$}$$
$endgroup$
Using polar coordinates:
$$r=frac{ep}{1-ecos theta}=frac{b^2}{a-ccos theta}$$
The extremities of the minor axis are:
$$(x,y)=(c,pm b)$$
Since
$$p=frac{b^2}{c}$$
The required locus is
$$fbox{$y^2=px$}$$
edited Dec 20 '18 at 13:47
answered Dec 20 '18 at 13:10
Ng Chung TakNg Chung Tak
14.7k31334
14.7k31334
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
|
show 1 more comment
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$ p= b^2/a $ Right?
$endgroup$
– Narasimham
Dec 20 '18 at 19:05
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
$p$ is not the semi-latus rectum $ell$ but the focus-directrix distance.
$endgroup$
– Ng Chung Tak
Dec 20 '18 at 19:28
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Can't it be solved by coordinate geometry rather than polar coordinates?
$endgroup$
– prashant sharma
Dec 21 '18 at 3:13
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
$begingroup$
Focus-origin conic looks simpler in polar coordinates than Cartesian. The picture shows the equation of this family of conics in Cartesian coordinates.
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 3:51
1
1
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
$begingroup$
@prashantsharma Using polar coordinates is coordinate geometry.
$endgroup$
– amd
Dec 21 '18 at 6:10
|
show 1 more comment
$begingroup$
I am trying a shortcut but stuck up..
Property of ellipse $ (x/a)^2+(y/b)^2 =1$ as sum of focal chord lengths is $2a$
$$ sqrt{x^2+y^2} + sqrt{(x-2c)^2+y^2} = 2 a = 2 sqrt{b^2+c^2} $$
Shifting origin to ellipse focus we can put
$$ c=x;, b=y $$
plug in and simplify the mix-up situation
$$ x^2-4cx+y^2 +y^2 = x^2+y^2 quadrightarrow y^2= 2 c x $$
If $p$ is directrix-focus distance Ng Chung Tak gets
$$ y^2= p x $$
I am immediately unable to figure out under what legitimate/interpretative conditions $p= 2c$ holds good,
$endgroup$
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
add a comment |
$begingroup$
I am trying a shortcut but stuck up..
Property of ellipse $ (x/a)^2+(y/b)^2 =1$ as sum of focal chord lengths is $2a$
$$ sqrt{x^2+y^2} + sqrt{(x-2c)^2+y^2} = 2 a = 2 sqrt{b^2+c^2} $$
Shifting origin to ellipse focus we can put
$$ c=x;, b=y $$
plug in and simplify the mix-up situation
$$ x^2-4cx+y^2 +y^2 = x^2+y^2 quadrightarrow y^2= 2 c x $$
If $p$ is directrix-focus distance Ng Chung Tak gets
$$ y^2= p x $$
I am immediately unable to figure out under what legitimate/interpretative conditions $p= 2c$ holds good,
$endgroup$
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
add a comment |
$begingroup$
I am trying a shortcut but stuck up..
Property of ellipse $ (x/a)^2+(y/b)^2 =1$ as sum of focal chord lengths is $2a$
$$ sqrt{x^2+y^2} + sqrt{(x-2c)^2+y^2} = 2 a = 2 sqrt{b^2+c^2} $$
Shifting origin to ellipse focus we can put
$$ c=x;, b=y $$
plug in and simplify the mix-up situation
$$ x^2-4cx+y^2 +y^2 = x^2+y^2 quadrightarrow y^2= 2 c x $$
If $p$ is directrix-focus distance Ng Chung Tak gets
$$ y^2= p x $$
I am immediately unable to figure out under what legitimate/interpretative conditions $p= 2c$ holds good,
$endgroup$
I am trying a shortcut but stuck up..
Property of ellipse $ (x/a)^2+(y/b)^2 =1$ as sum of focal chord lengths is $2a$
$$ sqrt{x^2+y^2} + sqrt{(x-2c)^2+y^2} = 2 a = 2 sqrt{b^2+c^2} $$
Shifting origin to ellipse focus we can put
$$ c=x;, b=y $$
plug in and simplify the mix-up situation
$$ x^2-4cx+y^2 +y^2 = x^2+y^2 quadrightarrow y^2= 2 c x $$
If $p$ is directrix-focus distance Ng Chung Tak gets
$$ y^2= p x $$
I am immediately unable to figure out under what legitimate/interpretative conditions $p= 2c$ holds good,
edited Dec 21 '18 at 6:39
answered Dec 21 '18 at 6:32
NarasimhamNarasimham
20.8k52158
20.8k52158
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
add a comment |
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
$begingroup$
For centre-origin ellipse, the directrices are $x=pm dfrac{a^2}{c}$ which varies with $e$, so $p=dfrac{a^2}{c}-c=dfrac{a^2-c^2}{c}=dfrac{b^2}{c}$
$endgroup$
– Ng Chung Tak
Dec 21 '18 at 8:39
add a comment |
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$begingroup$
By given point and "a" point I understood as a given fixed line and any line..Is that correct?
$endgroup$
– Narasimham
Dec 20 '18 at 18:47
$begingroup$
@Narasimham Line and point are fixed at a time.
$endgroup$
– prashant sharma
Dec 21 '18 at 3:11