Mixing
Common point between ellipse and tangent passing through external point
Clash Royale CLAN TAG#URR8PPP
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Given an ellipse $fracx^2a^2 + fracy^2b^2 = 1$ and a point $(u, v)$ not on the ellipse, I want to find two points that lie on the ellipse and on the two tangents of the ellipse passing through $(u, v)$.
Attempted solution:
Define $f(x, y) = fracx^2a^2 + fracy^2b^2 - 1$. A point $(x, y)$ is on the ellipse if $f(x, y) = 0$.
A normal of the ellipse at $(x, y)$ is $nabla f(x, y) = 2left(fracxa^2, fracyb^2right)$.
The point $(x, y)$ I am looking for has to satisfy two things: (i) it has to lie on the ellipse and (ii) the normal at $(x, y)$ has to be orthogonal to the line connecting $(x, y)$ and $(u, v)$:
(i): $fracx^2a^2 + fracy^2b^2 - 1 = 0$
(ii): $left(fracxa^2, fracyb^2right) cdot (x-u, y-v) = 0$.
Rewriting (ii), we get:
(ii): $fracx^2a^2 - fracuxa^2 + fracy^2b^2 - fracvyb^2 = 0$.
How to solve (i) and (ii) for $(x, y)$? From (i), we get $fracx^2a^2 + fracy^2b^2 = 1$, which we can plug into (ii) to get
(iii) $fracuxa^2 + fracvyb^2 = 1$.
But this is an equation of a whole line and not for two points. How to get $(x, y)$ from here? I could solve (iii) for $y$ and then plug the result in (i) to end up with one equation for $x$. But I used up (i) to get to (iii). I cannot go back to (i) from (iii), can I?
quadratics conic-sections
edited Jul 27 at 22:12
Math Lover
12.3k21232
asked Jul 27 at 21:36
chp
519
add a comment |Â
up vote
0
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Given an ellipse $fracx^2a^2 + fracy^2b^2 = 1$ and a point $(u, v)$ not on the ellipse, I want to find two points that lie on the ellipse and on the two tangents of the ellipse passing through $(u, v)$.
Attempted solution:
Define $f(x, y) = fracx^2a^2 + fracy^2b^2 - 1$. A point $(x, y)$ is on the ellipse if $f(x, y) = 0$.
A normal of the ellipse at $(x, y)$ is $nabla f(x, y) = 2left(fracxa^2, fracyb^2right)$.
The point $(x, y)$ I am looking for has to satisfy two things: (i) it has to lie on the ellipse and (ii) the normal at $(x, y)$ has to be orthogonal to the line connecting $(x, y)$ and $(u, v)$:
(i): $fracx^2a^2 + fracy^2b^2 - 1 = 0$
(ii): $left(fracxa^2, fracyb^2right) cdot (x-u, y-v) = 0$.
Rewriting (ii), we get:
(ii): $fracx^2a^2 - fracuxa^2 + fracy^2b^2 - fracvyb^2 = 0$.
How to solve (i) and (ii) for $(x, y)$? From (i), we get $fracx^2a^2 + fracy^2b^2 = 1$, which we can plug into (ii) to get
(iii) $fracuxa^2 + fracvyb^2 = 1$.
But this is an equation of a whole line and not for two points. How to get $(x, y)$ from here? I could solve (iii) for $y$ and then plug the result in (i) to end up with one equation for $x$. But I used up (i) to get to (iii). I cannot go back to (i) from (iii), can I?
quadratics conic-sections
edited Jul 27 at 22:12
Math Lover
12.3k21232
asked Jul 27 at 21:36
chp
519
Yes, you can...
– Aretino
Jul 27 at 21:39
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Given an ellipse $fracx^2a^2 + fracy^2b^2 = 1$ and a point $(u, v)$ not on the ellipse, I want to find two points that lie on the ellipse and on the two tangents of the ellipse passing through $(u, v)$.
Attempted solution:
Define $f(x, y) = fracx^2a^2 + fracy^2b^2 - 1$. A point $(x, y)$ is on the ellipse if $f(x, y) = 0$.
A normal of the ellipse at $(x, y)$ is $nabla f(x, y) = 2left(fracxa^2, fracyb^2right)$.
The point $(x, y)$ I am looking for has to satisfy two things: (i) it has to lie on the ellipse and (ii) the normal at $(x, y)$ has to be orthogonal to the line connecting $(x, y)$ and $(u, v)$:
(i): $fracx^2a^2 + fracy^2b^2 - 1 = 0$
(ii): $left(fracxa^2, fracyb^2right) cdot (x-u, y-v) = 0$.
Rewriting (ii), we get:
(ii): $fracx^2a^2 - fracuxa^2 + fracy^2b^2 - fracvyb^2 = 0$.
How to solve (i) and (ii) for $(x, y)$? From (i), we get $fracx^2a^2 + fracy^2b^2 = 1$, which we can plug into (ii) to get
(iii) $fracuxa^2 + fracvyb^2 = 1$.
But this is an equation of a whole line and not for two points. How to get $(x, y)$ from here? I could solve (iii) for $y$ and then plug the result in (i) to end up with one equation for $x$. But I used up (i) to get to (iii). I cannot go back to (i) from (iii), can I?
quadratics conic-sections
edited Jul 27 at 22:12
Math Lover
12.3k21232
asked Jul 27 at 21:36
chp
519
Given an ellipse $fracx^2a^2 + fracy^2b^2 = 1$ and a point $(u, v)$ not on the ellipse, I want to find two points that lie on the ellipse and on the two tangents of the ellipse passing through $(u, v)$.
Attempted solution:
Define $f(x, y) = fracx^2a^2 + fracy^2b^2 - 1$. A point $(x, y)$ is on the ellipse if $f(x, y) = 0$.
A normal of the ellipse at $(x, y)$ is $nabla f(x, y) = 2left(fracxa^2, fracyb^2right)$.
The point $(x, y)$ I am looking for has to satisfy two things: (i) it has to lie on the ellipse and (ii) the normal at $(x, y)$ has to be orthogonal to the line connecting $(x, y)$ and $(u, v)$:
(i): $fracx^2a^2 + fracy^2b^2 - 1 = 0$
(ii): $left(fracxa^2, fracyb^2right) cdot (x-u, y-v) = 0$.
Rewriting (ii), we get:
(ii): $fracx^2a^2 - fracuxa^2 + fracy^2b^2 - fracvyb^2 = 0$.
How to solve (i) and (ii) for $(x, y)$? From (i), we get $fracx^2a^2 + fracy^2b^2 = 1$, which we can plug into (ii) to get
(iii) $fracuxa^2 + fracvyb^2 = 1$.
But this is an equation of a whole line and not for two points. How to get $(x, y)$ from here? I could solve (iii) for $y$ and then plug the result in (i) to end up with one equation for $x$. But I used up (i) to get to (iii). I cannot go back to (i) from (iii), can I?
quadratics conic-sections
edited Jul 27 at 22:12
Math Lover
12.3k21232
asked Jul 27 at 21:36
chp
519
edited Jul 27 at 22:12
Math Lover
12.3k21232
edited Jul 27 at 22:12
Math Lover
12.3k21232
edited Jul 27 at 22:12
Math Lover
12.3k21232
12.3k21232
asked Jul 27 at 21:36
chp
519
asked Jul 27 at 21:36
chp
519
asked Jul 27 at 21:36
chp
519
519
Yes, you can...
– Aretino
Jul 27 at 21:39
add a comment |Â
Yes, you can...
– Aretino
Jul 27 at 21:39
Yes, you can...
– Aretino
Jul 27 at 21:39
Yes, you can...
– Aretino
Jul 27 at 21:39
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
A simple method:
Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.
Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation
$$y-v=t(x-u)iff y=t(x-u)+v qquad(tinbf R).$$
Replace $y$ in the equation of the ellipse, which you ma rewrite as
$$b^2x^2+a^2y^2=a^2b^2.$$
This yields the equation
$$b^2x^2+a^2bigl(t(x-u)+vbigl)^2=a^2b^2.$$
This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.
Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $;Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.
answered Jul 27 at 21:55
Bernard
110k635102
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
add a comment |Â
up vote
0
down vote
- $(u,v)$ is known as pole of the polar (chord), namely
$$fracuxa^2+fracvyb^2=1$$
- Substitute $y=dfracb^2vleft( 1-dfracuxa^2 right)$ into the ellipse:
$$fracx^2a^2+fracb^2v^2left( 1-fracuxa^2 right)^2=1$$
$$fracv^2x^2b^2+a^2left( 1-fracuxa^2 right)^2=fraca^2v^2b^2$$
$$left( fracu^2a^2+fracv^2b^2 right)x^2-
2ux+a^2left( 1-fracv^2b^2 right)=0$$
- Sum and product of roots:
$$x_1+x_2=frac2udfracu^2a^2+dfracv^2b^2$$
$$x_1 x_2=fraca^2left( 1-dfracv^2b^2 right)dfracu^2a^2+dfracv^2b^2$$
- Points of contact:
$$left(
fracu pm dfracavb sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 , , ,
fracv mp dfracbua sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 ,
right)$$
Standard results are summarized in the diagram below where $A(x',y')$ is the pole, $B(x_1,y_1)$ and $C(x_2,y_2)$ are the points of contact.
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
A simple method:
Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.
Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation
$$y-v=t(x-u)iff y=t(x-u)+v qquad(tinbf R).$$
Replace $y$ in the equation of the ellipse, which you ma rewrite as
$$b^2x^2+a^2y^2=a^2b^2.$$
This yields the equation
$$b^2x^2+a^2bigl(t(x-u)+vbigl)^2=a^2b^2.$$
This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.
Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $;Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.
answered Jul 27 at 21:55
Bernard
110k635102
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
add a comment |Â
up vote
1
down vote
A simple method:
Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.
Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation
$$y-v=t(x-u)iff y=t(x-u)+v qquad(tinbf R).$$
Replace $y$ in the equation of the ellipse, which you ma rewrite as
$$b^2x^2+a^2y^2=a^2b^2.$$
This yields the equation
$$b^2x^2+a^2bigl(t(x-u)+vbigl)^2=a^2b^2.$$
This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.
Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $;Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.
answered Jul 27 at 21:55
Bernard
110k635102
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A simple method:
Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.
Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation
$$y-v=t(x-u)iff y=t(x-u)+v qquad(tinbf R).$$
Replace $y$ in the equation of the ellipse, which you ma rewrite as
$$b^2x^2+a^2y^2=a^2b^2.$$
This yields the equation
$$b^2x^2+a^2bigl(t(x-u)+vbigl)^2=a^2b^2.$$
This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.
Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $;Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.
answered Jul 27 at 21:55
Bernard
110k635102
A simple method:
Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.
Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation
$$y-v=t(x-u)iff y=t(x-u)+v qquad(tinbf R).$$
Replace $y$ in the equation of the ellipse, which you ma rewrite as
$$b^2x^2+a^2y^2=a^2b^2.$$
This yields the equation
$$b^2x^2+a^2bigl(t(x-u)+vbigl)^2=a^2b^2.$$
This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.
Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $;Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.
answered Jul 27 at 21:55
Bernard
110k635102
answered Jul 27 at 21:55
Bernard
110k635102
answered Jul 27 at 21:55
Bernard
110k635102
answered Jul 27 at 21:55
Bernard
110k635102
110k635102
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
add a comment |Â
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
A very efficient way IMHO.
– Jyrki Lahtonen
Jul 27 at 22:09
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist.
– Doug M
Jul 27 at 22:30
add a comment |Â
up vote
0
down vote
- $(u,v)$ is known as pole of the polar (chord), namely
$$fracuxa^2+fracvyb^2=1$$
- Substitute $y=dfracb^2vleft( 1-dfracuxa^2 right)$ into the ellipse:
$$fracx^2a^2+fracb^2v^2left( 1-fracuxa^2 right)^2=1$$
$$fracv^2x^2b^2+a^2left( 1-fracuxa^2 right)^2=fraca^2v^2b^2$$
$$left( fracu^2a^2+fracv^2b^2 right)x^2-
2ux+a^2left( 1-fracv^2b^2 right)=0$$
- Sum and product of roots:
$$x_1+x_2=frac2udfracu^2a^2+dfracv^2b^2$$
$$x_1 x_2=fraca^2left( 1-dfracv^2b^2 right)dfracu^2a^2+dfracv^2b^2$$
- Points of contact:
$$left(
fracu pm dfracavb sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 , , ,
fracv mp dfracbua sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 ,
right)$$
Standard results are summarized in the diagram below where $A(x',y')$ is the pole, $B(x_1,y_1)$ and $C(x_2,y_2)$ are the points of contact.
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
add a comment |Â
up vote
0
down vote
- $(u,v)$ is known as pole of the polar (chord), namely
$$fracuxa^2+fracvyb^2=1$$
- Substitute $y=dfracb^2vleft( 1-dfracuxa^2 right)$ into the ellipse:
$$fracx^2a^2+fracb^2v^2left( 1-fracuxa^2 right)^2=1$$
$$fracv^2x^2b^2+a^2left( 1-fracuxa^2 right)^2=fraca^2v^2b^2$$
$$left( fracu^2a^2+fracv^2b^2 right)x^2-
2ux+a^2left( 1-fracv^2b^2 right)=0$$
- Sum and product of roots:
$$x_1+x_2=frac2udfracu^2a^2+dfracv^2b^2$$
$$x_1 x_2=fraca^2left( 1-dfracv^2b^2 right)dfracu^2a^2+dfracv^2b^2$$
- Points of contact:
$$left(
fracu pm dfracavb sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 , , ,
fracv mp dfracbua sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 ,
right)$$
Standard results are summarized in the diagram below where $A(x',y')$ is the pole, $B(x_1,y_1)$ and $C(x_2,y_2)$ are the points of contact.
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- $(u,v)$ is known as pole of the polar (chord), namely
$$fracuxa^2+fracvyb^2=1$$
- Substitute $y=dfracb^2vleft( 1-dfracuxa^2 right)$ into the ellipse:
$$fracx^2a^2+fracb^2v^2left( 1-fracuxa^2 right)^2=1$$
$$fracv^2x^2b^2+a^2left( 1-fracuxa^2 right)^2=fraca^2v^2b^2$$
$$left( fracu^2a^2+fracv^2b^2 right)x^2-
2ux+a^2left( 1-fracv^2b^2 right)=0$$
- Sum and product of roots:
$$x_1+x_2=frac2udfracu^2a^2+dfracv^2b^2$$
$$x_1 x_2=fraca^2left( 1-dfracv^2b^2 right)dfracu^2a^2+dfracv^2b^2$$
- Points of contact:
$$left(
fracu pm dfracavb sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 , , ,
fracv mp dfracbua sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 ,
right)$$
Standard results are summarized in the diagram below where $A(x',y')$ is the pole, $B(x_1,y_1)$ and $C(x_2,y_2)$ are the points of contact.
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
- $(u,v)$ is known as pole of the polar (chord), namely
$$fracuxa^2+fracvyb^2=1$$
- Substitute $y=dfracb^2vleft( 1-dfracuxa^2 right)$ into the ellipse:
$$fracx^2a^2+fracb^2v^2left( 1-fracuxa^2 right)^2=1$$
$$fracv^2x^2b^2+a^2left( 1-fracuxa^2 right)^2=fraca^2v^2b^2$$
$$left( fracu^2a^2+fracv^2b^2 right)x^2-
2ux+a^2left( 1-fracv^2b^2 right)=0$$
- Sum and product of roots:
$$x_1+x_2=frac2udfracu^2a^2+dfracv^2b^2$$
$$x_1 x_2=fraca^2left( 1-dfracv^2b^2 right)dfracu^2a^2+dfracv^2b^2$$
- Points of contact:
$$left(
fracu pm dfracavb sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 , , ,
fracv mp dfracbua sqrtdfracu^2a^2+dfracv^2b^2-1
dfracu^2a^2+dfracv^2b^2 ,
right)$$
Standard results are summarized in the diagram below where $A(x',y')$ is the pole, $B(x_1,y_1)$ and $C(x_2,y_2)$ are the points of contact.
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
edited Jul 28 at 8:55
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
answered Jul 28 at 8:26
Ng Chung Tak
12.9k31129
12.9k31129
add a comment |Â
add a comment |Â
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Yes, you can...
– Aretino
Jul 27 at 21:39