Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to...
Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex
I've been trying to prove this by plugging in the negative reciprocal of the slope of the tangent at a point $(x, y)$ into a line which passes through that point and the axis of symmetry. Then I plug the value of the focus into the result and solve for $x$. However the slope is undefined for any line parallel to the axis of symmetry.
calculus conic-sections tangent-line
edited 2 days ago
Blue
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
add a comment |
Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex
I've been trying to prove this by plugging in the negative reciprocal of the slope of the tangent at a point $(x, y)$ into a line which passes through that point and the axis of symmetry. Then I plug the value of the focus into the result and solve for $x$. However the slope is undefined for any line parallel to the axis of symmetry.
calculus conic-sections tangent-line
edited 2 days ago
Blue
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
add a comment |
Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex
I've been trying to prove this by plugging in the negative reciprocal of the slope of the tangent at a point $(x, y)$ into a line which passes through that point and the axis of symmetry. Then I plug the value of the focus into the result and solve for $x$. However the slope is undefined for any line parallel to the axis of symmetry.
calculus conic-sections tangent-line
edited 2 days ago
Blue
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex
I've been trying to prove this by plugging in the negative reciprocal of the slope of the tangent at a point $(x, y)$ into a line which passes through that point and the axis of symmetry. Then I plug the value of the focus into the result and solve for $x$. However the slope is undefined for any line parallel to the axis of symmetry.
calculus conic-sections tangent-line
calculus conic-sections tangent-line
edited 2 days ago
Blue
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
edited 2 days ago
Blue
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
edited 2 days ago
Blue
47.7k870151
edited 2 days ago
Blue
47.7k870151
edited 2 days ago
Blue
47.7k870151
47.7k870151
asked Feb 6 '14 at 9:41
Tessera
175
asked Feb 6 '14 at 9:41
Tessera
175
asked Feb 6 '14 at 9:41
Tessera
175
175
add a comment |
add a comment |
2 Answers
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Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
answered 2 days ago
Aretino
22.7k21442
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Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,frac{c^2}{4a})$ is $frac{c}{2a}$ and the tangent equation is $$y-frac{c^2}{4a}=frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum.
$$d=frac{4a(a)-2c(0)-c^2+2c^2}{sqrt{16a^2+4c^2}}
\=frac{4a^2+c^2}{sqrt{16a^2+4c^2}}
\=frac{1}{2}sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
answered Feb 6 '14 at 9:54
Semsem
6,51531534
add a comment |
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2 Answers
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2 Answers
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Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
answered 2 days ago
Aretino
22.7k21442
add a comment |
Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
answered 2 days ago
Aretino
22.7k21442
add a comment |
Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
answered 2 days ago
Aretino
22.7k21442
Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
answered 2 days ago
Aretino
22.7k21442
edited 2 days ago
answered 2 days ago
Aretino
22.7k21442
answered 2 days ago
Aretino
22.7k21442
answered 2 days ago
Aretino
22.7k21442
22.7k21442
add a comment |
add a comment |
Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,frac{c^2}{4a})$ is $frac{c}{2a}$ and the tangent equation is $$y-frac{c^2}{4a}=frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum.
$$d=frac{4a(a)-2c(0)-c^2+2c^2}{sqrt{16a^2+4c^2}}
\=frac{4a^2+c^2}{sqrt{16a^2+4c^2}}
\=frac{1}{2}sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
answered Feb 6 '14 at 9:54
Semsem
6,51531534
add a comment |
Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,frac{c^2}{4a})$ is $frac{c}{2a}$ and the tangent equation is $$y-frac{c^2}{4a}=frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum.
$$d=frac{4a(a)-2c(0)-c^2+2c^2}{sqrt{16a^2+4c^2}}
\=frac{4a^2+c^2}{sqrt{16a^2+4c^2}}
\=frac{1}{2}sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
answered Feb 6 '14 at 9:54
Semsem
6,51531534
add a comment |
Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,frac{c^2}{4a})$ is $frac{c}{2a}$ and the tangent equation is $$y-frac{c^2}{4a}=frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum.
$$d=frac{4a(a)-2c(0)-c^2+2c^2}{sqrt{16a^2+4c^2}}
\=frac{4a^2+c^2}{sqrt{16a^2+4c^2}}
\=frac{1}{2}sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
answered Feb 6 '14 at 9:54
Semsem
6,51531534
Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,frac{c^2}{4a})$ is $frac{c}{2a}$ and the tangent equation is $$y-frac{c^2}{4a}=frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum.
$$d=frac{4a(a)-2c(0)-c^2+2c^2}{sqrt{16a^2+4c^2}}
\=frac{4a^2+c^2}{sqrt{16a^2+4c^2}}
\=frac{1}{2}sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
answered Feb 6 '14 at 9:54
Semsem
6,51531534
edited Feb 6 '14 at 10:06
answered Feb 6 '14 at 9:54
Semsem
6,51531534
answered Feb 6 '14 at 9:54
Semsem
6,51531534
answered Feb 6 '14 at 9:54
Semsem
6,51531534
6,51531534
add a comment |
add a comment |
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