Proving a point lies on a ellipse
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Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
geometry conic-sections
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
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up vote
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Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
geometry conic-sections
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
307
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
geometry conic-sections
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
307
Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
geometry conic-sections
geometry conic-sections
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
307
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
307
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
edited Nov 22 at 11:48
Ng Chung Tak
13.8k31234
13.8k31234
asked Nov 22 at 11:15
Matlab rookie
307
asked Nov 22 at 11:15
Matlab rookie
307
asked Nov 22 at 11:15
Matlab rookie
307
307
add a comment |
add a comment |
1 Answer
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Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$frac{X-x}X=frac{b}{a+b}$$
$$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$frac{Y-y}{Y}=frac{a}{a+b}$$
$$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$
Now,
$$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$frac{X-x}X=frac{b}{a+b}$$
$$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$frac{Y-y}{Y}=frac{a}{a+b}$$
$$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$
Now,
$$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
add a comment |
up vote
1
down vote
accepted
Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$frac{X-x}X=frac{b}{a+b}$$
$$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$frac{Y-y}{Y}=frac{a}{a+b}$$
$$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$
Now,
$$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$frac{X-x}X=frac{b}{a+b}$$
$$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$frac{Y-y}{Y}=frac{a}{a+b}$$
$$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$
Now,
$$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$frac{X-x}X=frac{b}{a+b}$$
$$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$frac{Y-y}{Y}=frac{a}{a+b}$$
$$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$
Now,
$$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
edited Nov 22 at 11:41
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
answered Nov 22 at 11:34
Jean-Claude Arbaut
14.7k63363
14.7k63363
add a comment |
add a comment |
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