A tangent to an ellipse makes angles $alpha$ with major axis and $beta$ with a focal radius; show that the eccentricity is $cosbeta/cosalpha$.

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If the tangent at any point of the ellipse make an angle $alpha$ with the major axis and an angle $beta$ with the focal radius of the point of contact, then show that the eccentricity of the ellipse is given by $$e=dfraccosbetacosalpha$$

How is this derived? Please explain with a proper diagram.

conic-sections

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edited Jul 27 at 18:57

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up vote
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If the tangent at any point of the ellipse make an angle $alpha$ with the major axis and an angle $beta$ with the focal radius of the point of contact, then show that the eccentricity of the ellipse is given by $$e=dfraccosbetacosalpha$$

How is this derived? Please explain with a proper diagram.

conic-sections

share|cite|improve this question

edited Jul 27 at 18:57

Blue

43.6k868141

asked Jul 27 at 4:00

Kaushal Saluja

1

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

If the tangent at any point of the ellipse make an angle $alpha$ with the major axis and an angle $beta$ with the focal radius of the point of contact, then show that the eccentricity of the ellipse is given by $$e=dfraccosbetacosalpha$$

How is this derived? Please explain with a proper diagram.

conic-sections

share|cite|improve this question

edited Jul 27 at 18:57

Blue

43.6k868141

asked Jul 27 at 4:00

Kaushal Saluja

1

If the tangent at any point of the ellipse make an angle $alpha$ with the major axis and an angle $beta$ with the focal radius of the point of contact, then show that the eccentricity of the ellipse is given by $$e=dfraccosbetacosalpha$$

How is this derived? Please explain with a proper diagram.

conic-sections

share|cite|improve this question

edited Jul 27 at 18:57

Blue

43.6k868141

asked Jul 27 at 4:00

Kaushal Saluja

1

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 18:57

Blue

43.6k868141

edited Jul 27 at 18:57

Blue

43.6k868141

edited Jul 27 at 18:57

Blue

43.6k868141

43.6k868141

asked Jul 27 at 4:00

Kaushal Saluja

1

asked Jul 27 at 4:00

Kaushal Saluja

1

asked Jul 27 at 4:00

Kaushal Saluja

1

1

add a comment | 

add a comment | 

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Let $F$ and $G$ be the foci of the ellipse.
The bisector $PB$ of $angle FPG$ is perpendicular to tangent $PQ$. Hence:
$$
beginalign
e&=FB+GBover FP+GPquadtext(definition)\
&=FBover FPquadtext(angle bisector theorem)\
&=sin(pi/2-beta)oversin(pi/2-alpha)quadtext(sine rule)\
&=cosbetaovercosalpha.
endalign
$$

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edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

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Let $F$ and $G$ be the foci of the ellipse.
The bisector $PB$ of $angle FPG$ is perpendicular to tangent $PQ$. Hence:
$$
beginalign
e&=FB+GBover FP+GPquadtext(definition)\
&=FBover FPquadtext(angle bisector theorem)\
&=sin(pi/2-beta)oversin(pi/2-alpha)quadtext(sine rule)\
&=cosbetaovercosalpha.
endalign
$$

share|cite|improve this answer

edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

21.7k21342

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up vote
1
down vote

Let $F$ and $G$ be the foci of the ellipse.
The bisector $PB$ of $angle FPG$ is perpendicular to tangent $PQ$. Hence:
$$
beginalign
e&=FB+GBover FP+GPquadtext(definition)\
&=FBover FPquadtext(angle bisector theorem)\
&=sin(pi/2-beta)oversin(pi/2-alpha)quadtext(sine rule)\
&=cosbetaovercosalpha.
endalign
$$

share|cite|improve this answer

edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

21.7k21342

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $F$ and $G$ be the foci of the ellipse.
The bisector $PB$ of $angle FPG$ is perpendicular to tangent $PQ$. Hence:
$$
beginalign
e&=FB+GBover FP+GPquadtext(definition)\
&=FBover FPquadtext(angle bisector theorem)\
&=sin(pi/2-beta)oversin(pi/2-alpha)quadtext(sine rule)\
&=cosbetaovercosalpha.
endalign
$$

share|cite|improve this answer

edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

21.7k21342

Let $F$ and $G$ be the foci of the ellipse.
The bisector $PB$ of $angle FPG$ is perpendicular to tangent $PQ$. Hence:
$$
beginalign
e&=FB+GBover FP+GPquadtext(definition)\
&=FBover FPquadtext(angle bisector theorem)\
&=sin(pi/2-beta)oversin(pi/2-alpha)quadtext(sine rule)\
&=cosbetaovercosalpha.
endalign
$$

share|cite|improve this answer

edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

21.7k21342

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 17:12

edited Aug 1 at 17:12

edited Aug 1 at 17:12

answered Jul 27 at 17:35

Aretino

21.7k21342

answered Jul 27 at 17:35

Aretino

21.7k21342

answered Jul 27 at 17:35

Aretino

21.7k21342

21.7k21342

add a comment | 

add a comment | 

 

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