Locus of points with parabola
$begingroup$
On parabola $y^2=2px$ at point $A$, a line $L_1$ passes that is
tangent to the parabola and cuts $x$ axis at point $B$. From $A$, a
line $L_2$ passes that is perpendicular to $x$ axis and cuts the
parabola at point $C$.
A line $L_3$ passes point $B$ and is perpendicular to $x$ axis. A line
$L_4$ passes point C and is parallel to $x$ axis.
Find the set of points $F$ where $L_3$ and $L_4$ intersect.
Book's answer is $y^2=-2px$
Graphic representation as I see it
My attempt to solve it
My intention was to present point $A$ as $(T,K)$ and eventually express $L_3$ and $L_4$ that way, find the expression of $T$ and $K$, and place them in the parabola's equation for the answer. Although I feel like the answer is there, I got lost. Would appreciate help
analytic-geometry locus
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
$endgroup$
add a comment |
$begingroup$
On parabola $y^2=2px$ at point $A$, a line $L_1$ passes that is
tangent to the parabola and cuts $x$ axis at point $B$. From $A$, a
line $L_2$ passes that is perpendicular to $x$ axis and cuts the
parabola at point $C$.
A line $L_3$ passes point $B$ and is perpendicular to $x$ axis. A line
$L_4$ passes point C and is parallel to $x$ axis.
Find the set of points $F$ where $L_3$ and $L_4$ intersect.
Book's answer is $y^2=-2px$
Graphic representation as I see it
My attempt to solve it
My intention was to present point $A$ as $(T,K)$ and eventually express $L_3$ and $L_4$ that way, find the expression of $T$ and $K$, and place them in the parabola's equation for the answer. Although I feel like the answer is there, I got lost. Would appreciate help
analytic-geometry locus
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
$endgroup$
add a comment |
$begingroup$
On parabola $y^2=2px$ at point $A$, a line $L_1$ passes that is
tangent to the parabola and cuts $x$ axis at point $B$. From $A$, a
line $L_2$ passes that is perpendicular to $x$ axis and cuts the
parabola at point $C$.
A line $L_3$ passes point $B$ and is perpendicular to $x$ axis. A line
$L_4$ passes point C and is parallel to $x$ axis.
Find the set of points $F$ where $L_3$ and $L_4$ intersect.
Book's answer is $y^2=-2px$
Graphic representation as I see it
My attempt to solve it
My intention was to present point $A$ as $(T,K)$ and eventually express $L_3$ and $L_4$ that way, find the expression of $T$ and $K$, and place them in the parabola's equation for the answer. Although I feel like the answer is there, I got lost. Would appreciate help
analytic-geometry locus
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
$endgroup$
On parabola $y^2=2px$ at point $A$, a line $L_1$ passes that is
tangent to the parabola and cuts $x$ axis at point $B$. From $A$, a
line $L_2$ passes that is perpendicular to $x$ axis and cuts the
parabola at point $C$.
A line $L_3$ passes point $B$ and is perpendicular to $x$ axis. A line
$L_4$ passes point C and is parallel to $x$ axis.
Find the set of points $F$ where $L_3$ and $L_4$ intersect.
Book's answer is $y^2=-2px$
Graphic representation as I see it
My attempt to solve it
My intention was to present point $A$ as $(T,K)$ and eventually express $L_3$ and $L_4$ that way, find the expression of $T$ and $K$, and place them in the parabola's equation for the answer. Although I feel like the answer is there, I got lost. Would appreciate help
analytic-geometry locus
analytic-geometry locus
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
edited Dec 19 '18 at 17:39
Shubham Johri
5,192717
5,192717
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
asked Dec 19 '18 at 16:59
Avi KenigAvi Kenig
34
34
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $A(x_0,y_0)$. The equation of $L_2$ is $L_2:x=x_0$. Since the parabola is symmetric about the $x$ axis, $L_2$ intersects the parabola again at $A'(x_0,-y_0)$. Thus, $L_4:y=-y_0$.
$y^2=2pximplies 2yy'=2pimplies y'=p/y$.
The equation of the tangent at $A(x_0,y_0)$ is given by $L_1:displaystylefrac{y-y_0}{x-x_0}=frac p{y_0}$.
The $x$ intercept of the tangent $L_1$ is $displaystyle x_0-frac{y_0^2}p=-x_0 because y_0^2=2px_0$. Thus, $L_3: x=-x_0$.
The intersection of $L_3,L_4$ is the point $(-x_0,-y_0)=(x,y)$. Since $(x_0,y_0)$ lies on the parabola, the required locus is:
$$y^2=-2px$$
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Let $A(x_0,y_0)$. The equation of $L_2$ is $L_2:x=x_0$. Since the parabola is symmetric about the $x$ axis, $L_2$ intersects the parabola again at $A'(x_0,-y_0)$. Thus, $L_4:y=-y_0$.
$y^2=2pximplies 2yy'=2pimplies y'=p/y$.
The equation of the tangent at $A(x_0,y_0)$ is given by $L_1:displaystylefrac{y-y_0}{x-x_0}=frac p{y_0}$.
The $x$ intercept of the tangent $L_1$ is $displaystyle x_0-frac{y_0^2}p=-x_0 because y_0^2=2px_0$. Thus, $L_3: x=-x_0$.
The intersection of $L_3,L_4$ is the point $(-x_0,-y_0)=(x,y)$. Since $(x_0,y_0)$ lies on the parabola, the required locus is:
$$y^2=-2px$$
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Let $A(x_0,y_0)$. The equation of $L_2$ is $L_2:x=x_0$. Since the parabola is symmetric about the $x$ axis, $L_2$ intersects the parabola again at $A'(x_0,-y_0)$. Thus, $L_4:y=-y_0$.
$y^2=2pximplies 2yy'=2pimplies y'=p/y$.
The equation of the tangent at $A(x_0,y_0)$ is given by $L_1:displaystylefrac{y-y_0}{x-x_0}=frac p{y_0}$.
The $x$ intercept of the tangent $L_1$ is $displaystyle x_0-frac{y_0^2}p=-x_0 because y_0^2=2px_0$. Thus, $L_3: x=-x_0$.
The intersection of $L_3,L_4$ is the point $(-x_0,-y_0)=(x,y)$. Since $(x_0,y_0)$ lies on the parabola, the required locus is:
$$y^2=-2px$$
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
Let $A(x_0,y_0)$. The equation of $L_2$ is $L_2:x=x_0$. Since the parabola is symmetric about the $x$ axis, $L_2$ intersects the parabola again at $A'(x_0,-y_0)$. Thus, $L_4:y=-y_0$.
$y^2=2pximplies 2yy'=2pimplies y'=p/y$.
The equation of the tangent at $A(x_0,y_0)$ is given by $L_1:displaystylefrac{y-y_0}{x-x_0}=frac p{y_0}$.
The $x$ intercept of the tangent $L_1$ is $displaystyle x_0-frac{y_0^2}p=-x_0 because y_0^2=2px_0$. Thus, $L_3: x=-x_0$.
The intersection of $L_3,L_4$ is the point $(-x_0,-y_0)=(x,y)$. Since $(x_0,y_0)$ lies on the parabola, the required locus is:
$$y^2=-2px$$
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
$endgroup$
Let $A(x_0,y_0)$. The equation of $L_2$ is $L_2:x=x_0$. Since the parabola is symmetric about the $x$ axis, $L_2$ intersects the parabola again at $A'(x_0,-y_0)$. Thus, $L_4:y=-y_0$.
$y^2=2pximplies 2yy'=2pimplies y'=p/y$.
The equation of the tangent at $A(x_0,y_0)$ is given by $L_1:displaystylefrac{y-y_0}{x-x_0}=frac p{y_0}$.
The $x$ intercept of the tangent $L_1$ is $displaystyle x_0-frac{y_0^2}p=-x_0 because y_0^2=2px_0$. Thus, $L_3: x=-x_0$.
The intersection of $L_3,L_4$ is the point $(-x_0,-y_0)=(x,y)$. Since $(x_0,y_0)$ lies on the parabola, the required locus is:
$$y^2=-2px$$
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
edited Dec 19 '18 at 18:04
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
answered Dec 19 '18 at 17:58
Shubham JohriShubham Johri
5,192717
5,192717
add a comment |
add a comment |
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