Coordinate Geometry: Locus












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 13:26









jayant98

655318




655318










asked Dec 20 '18 at 9:49









prashant sharmaprashant sharma

797




797












  • $begingroup$
    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










  • $begingroup$
    @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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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



$$fbox{$y^2=px$}$$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ 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



















0












$begingroup$

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,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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



$$fbox{$y^2=px$}$$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ 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
















4












$begingroup$

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



$$fbox{$y^2=px$}$$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ 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














4












4








4





$begingroup$

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



$$fbox{$y^2=px$}$$



enter image description here






share|cite|improve this answer











$endgroup$



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



$$fbox{$y^2=px$}$$



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    $ 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











0












$begingroup$

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,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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,






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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,






share|cite|improve this answer











$endgroup$



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,







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 6:39

























answered Dec 21 '18 at 6:32









NarasimhamNarasimham

20.8k62158




20.8k62158












  • $begingroup$
    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






$begingroup$
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




















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