Mixing
Verifying the vertices of a Hyperbola
$begingroup$
In this exercise it is required to verify that the points O & A(4,0)
are the vertices of the Hyperbola H , as you can see in the marked part.
Can someone help verify that ?
conic-sections hyperbolic-geometry
edited Jan 19 at 9:24
Glorfindel
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
$endgroup$
add a comment |
$begingroup$
In this exercise it is required to verify that the points O & A(4,0)
are the vertices of the Hyperbola H , as you can see in the marked part.
Can someone help verify that ?
conic-sections hyperbolic-geometry
edited Jan 19 at 9:24
Glorfindel
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
$endgroup$
add a comment |
$begingroup$
In this exercise it is required to verify that the points O & A(4,0)
are the vertices of the Hyperbola H , as you can see in the marked part.
Can someone help verify that ?
conic-sections hyperbolic-geometry
edited Jan 19 at 9:24
Glorfindel
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
$endgroup$
In this exercise it is required to verify that the points O & A(4,0)
are the vertices of the Hyperbola H , as you can see in the marked part.
Can someone help verify that ?
conic-sections hyperbolic-geometry
conic-sections hyperbolic-geometry
edited Jan 19 at 9:24
Glorfindel
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
edited Jan 19 at 9:24
Glorfindel
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
edited Jan 19 at 9:24
Glorfindel
3,41981830
edited Jan 19 at 9:24
Glorfindel
3,41981830
edited Jan 19 at 9:24
Glorfindel
3,41981830
3,41981830
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
asked Jun 22 '13 at 14:46
A.JouniA.Jouni
1249
1249
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Some ideas:
If $,M=(h,k);$ , then it must be that $,MM'=|h-1|=r;,;r=$ the circle's radius.
We also have, since the triangle $,Delta TFM;$ is a $,30^circ-60^circ-90^circ;$ triangle, that
$$MF=2r;,;FT=sqrt 3,r$$
and the above already proves $;(1);$ . Now doing a little analytic geometry:
$$4=frac{MF^2}{MM'^2}=frac{(h+2)^2+k^2}{r^2}implies (h+2)^2+k^2=4(h-1)^2implies$$
$$h^2+k^2+4h+4=4h^2-8h+4implies 3(h^2-4h)-k^2=0implies$$
$$3left(h-2right)^2-k^2=12implies frac{(h-2)^2}{2^2}-frac{k^2}{(2sqrt3)^2}=1$$
and we have the standard equation of a hyperbola.
Try now to complete your question's answer.
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some ideas:
If $,M=(h,k);$ , then it must be that $,MM'=|h-1|=r;,;r=$ the circle's radius.
We also have, since the triangle $,Delta TFM;$ is a $,30^circ-60^circ-90^circ;$ triangle, that
$$MF=2r;,;FT=sqrt 3,r$$
and the above already proves $;(1);$ . Now doing a little analytic geometry:
$$4=frac{MF^2}{MM'^2}=frac{(h+2)^2+k^2}{r^2}implies (h+2)^2+k^2=4(h-1)^2implies$$
$$h^2+k^2+4h+4=4h^2-8h+4implies 3(h^2-4h)-k^2=0implies$$
$$3left(h-2right)^2-k^2=12implies frac{(h-2)^2}{2^2}-frac{k^2}{(2sqrt3)^2}=1$$
and we have the standard equation of a hyperbola.
Try now to complete your question's answer.
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Some ideas:
If $,M=(h,k);$ , then it must be that $,MM'=|h-1|=r;,;r=$ the circle's radius.
We also have, since the triangle $,Delta TFM;$ is a $,30^circ-60^circ-90^circ;$ triangle, that
$$MF=2r;,;FT=sqrt 3,r$$
and the above already proves $;(1);$ . Now doing a little analytic geometry:
$$4=frac{MF^2}{MM'^2}=frac{(h+2)^2+k^2}{r^2}implies (h+2)^2+k^2=4(h-1)^2implies$$
$$h^2+k^2+4h+4=4h^2-8h+4implies 3(h^2-4h)-k^2=0implies$$
$$3left(h-2right)^2-k^2=12implies frac{(h-2)^2}{2^2}-frac{k^2}{(2sqrt3)^2}=1$$
and we have the standard equation of a hyperbola.
Try now to complete your question's answer.
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Some ideas:
If $,M=(h,k);$ , then it must be that $,MM'=|h-1|=r;,;r=$ the circle's radius.
We also have, since the triangle $,Delta TFM;$ is a $,30^circ-60^circ-90^circ;$ triangle, that
$$MF=2r;,;FT=sqrt 3,r$$
and the above already proves $;(1);$ . Now doing a little analytic geometry:
$$4=frac{MF^2}{MM'^2}=frac{(h+2)^2+k^2}{r^2}implies (h+2)^2+k^2=4(h-1)^2implies$$
$$h^2+k^2+4h+4=4h^2-8h+4implies 3(h^2-4h)-k^2=0implies$$
$$3left(h-2right)^2-k^2=12implies frac{(h-2)^2}{2^2}-frac{k^2}{(2sqrt3)^2}=1$$
and we have the standard equation of a hyperbola.
Try now to complete your question's answer.
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
$endgroup$
Some ideas:
If $,M=(h,k);$ , then it must be that $,MM'=|h-1|=r;,;r=$ the circle's radius.
We also have, since the triangle $,Delta TFM;$ is a $,30^circ-60^circ-90^circ;$ triangle, that
$$MF=2r;,;FT=sqrt 3,r$$
and the above already proves $;(1);$ . Now doing a little analytic geometry:
$$4=frac{MF^2}{MM'^2}=frac{(h+2)^2+k^2}{r^2}implies (h+2)^2+k^2=4(h-1)^2implies$$
$$h^2+k^2+4h+4=4h^2-8h+4implies 3(h^2-4h)-k^2=0implies$$
$$3left(h-2right)^2-k^2=12implies frac{(h-2)^2}{2^2}-frac{k^2}{(2sqrt3)^2}=1$$
and we have the standard equation of a hyperbola.
Try now to complete your question's answer.
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
answered Jun 22 '13 at 19:01
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
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