Mixing
Foot of perpendicular on a chord of a conic
$begingroup$
For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:
$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$
By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.
Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.
analytic-geometry conic-sections
edited Nov 14 '15 at 7:17
G-man
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
$endgroup$
add a comment |
$begingroup$
For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:
$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$
By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.
Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.
analytic-geometry conic-sections
edited Nov 14 '15 at 7:17
G-man
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
$endgroup$
$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06
add a comment |
$begingroup$
For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:
$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$
By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.
Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.
analytic-geometry conic-sections
edited Nov 14 '15 at 7:17
G-man
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
$endgroup$
For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:
$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$
By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.
Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.
analytic-geometry conic-sections
analytic-geometry conic-sections
edited Nov 14 '15 at 7:17
G-man
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
edited Nov 14 '15 at 7:17
G-man
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
edited Nov 14 '15 at 7:17
G-man
4,49331344
edited Nov 14 '15 at 7:17
G-man
4,49331344
edited Nov 14 '15 at 7:17
G-man
4,49331344
4,49331344
asked Nov 14 '15 at 7:03
Sat DSat D
423212
asked Nov 14 '15 at 7:03
Sat DSat D
423212
asked Nov 14 '15 at 7:03
Sat DSat D
423212
423212
$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06
add a comment |
$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06
$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06
$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.
answered Nov 17 '15 at 7:50
Sat DSat D
423212
$endgroup$
add a comment |
$begingroup$
Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
$$
A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
$$
A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
$$
B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
$$
You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.
answered Nov 17 '15 at 7:50
Sat DSat D
423212
$endgroup$
add a comment |
$begingroup$
Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.
answered Nov 17 '15 at 7:50
Sat DSat D
423212
$endgroup$
add a comment |
$begingroup$
Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.
answered Nov 17 '15 at 7:50
Sat DSat D
423212
$endgroup$
Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.
answered Nov 17 '15 at 7:50
Sat DSat D
423212
edited Jan 26 '16 at 5:19
answered Nov 17 '15 at 7:50
Sat DSat D
423212
answered Nov 17 '15 at 7:50
Sat DSat D
423212
answered Nov 17 '15 at 7:50
Sat DSat D
423212
423212
add a comment |
add a comment |
$begingroup$
Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
$$
A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
$$
A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
$$
B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
$$
You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
$endgroup$
add a comment |
$begingroup$
Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
$$
A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
$$
A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
$$
B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
$$
You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
$endgroup$
add a comment |
$begingroup$
Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
$$
A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
$$
A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
$$
B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
$$
You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
$endgroup$
Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
$$
A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
$$
A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
$$
B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
$$
You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
answered Jan 7 at 18:59
AretinoAretino
23.1k21443
23.1k21443
add a comment |
add a comment |
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$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06