Minimum distance of curve from origin
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I have a parabola $(y+5)^2 = 4x$ and I need to find its minimum distance from origin. Scientific calculators aren't allowed.
I have tried :
1) Substituting parametric coordinates $(rcos Q, rsin Q)$ but the expression of $r$ in $Q$ obtained after differenciation doesn't give roots which can be calculated by hand.
2) Used point form of tangent for $(x_1,y_1)$ on parabola and made line joining origin and $(x_1,y_1)$ perpendicular but it forms a cubic equation which again can't be solved by hand.
Help please!
conic-sections coordinate-systems parametric maxima-minima
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
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add a comment |
$begingroup$
I have a parabola $(y+5)^2 = 4x$ and I need to find its minimum distance from origin. Scientific calculators aren't allowed.
I have tried :
1) Substituting parametric coordinates $(rcos Q, rsin Q)$ but the expression of $r$ in $Q$ obtained after differenciation doesn't give roots which can be calculated by hand.
2) Used point form of tangent for $(x_1,y_1)$ on parabola and made line joining origin and $(x_1,y_1)$ perpendicular but it forms a cubic equation which again can't be solved by hand.
Help please!
conic-sections coordinate-systems parametric maxima-minima
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
$endgroup$
add a comment |
$begingroup$
I have a parabola $(y+5)^2 = 4x$ and I need to find its minimum distance from origin. Scientific calculators aren't allowed.
I have tried :
1) Substituting parametric coordinates $(rcos Q, rsin Q)$ but the expression of $r$ in $Q$ obtained after differenciation doesn't give roots which can be calculated by hand.
2) Used point form of tangent for $(x_1,y_1)$ on parabola and made line joining origin and $(x_1,y_1)$ perpendicular but it forms a cubic equation which again can't be solved by hand.
Help please!
conic-sections coordinate-systems parametric maxima-minima
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
$endgroup$
I have a parabola $(y+5)^2 = 4x$ and I need to find its minimum distance from origin. Scientific calculators aren't allowed.
I have tried :
1) Substituting parametric coordinates $(rcos Q, rsin Q)$ but the expression of $r$ in $Q$ obtained after differenciation doesn't give roots which can be calculated by hand.
2) Used point form of tangent for $(x_1,y_1)$ on parabola and made line joining origin and $(x_1,y_1)$ perpendicular but it forms a cubic equation which again can't be solved by hand.
Help please!
conic-sections coordinate-systems parametric maxima-minima
conic-sections coordinate-systems parametric maxima-minima
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
edited Dec 6 '18 at 5:18
Thomas Shelby
2,765421
2,765421
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
asked Dec 6 '18 at 4:47
Meet ShahMeet Shah
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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The solution DO involves solving a cubic equation, which, by the way, CAN be solved by hand with Cardano's method.
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
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$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
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– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
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
$begingroup$
The solution DO involves solving a cubic equation, which, by the way, CAN be solved by hand with Cardano's method.
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
$endgroup$
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
add a comment |
$begingroup$
The solution DO involves solving a cubic equation, which, by the way, CAN be solved by hand with Cardano's method.
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
$endgroup$
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
add a comment |
$begingroup$
The solution DO involves solving a cubic equation, which, by the way, CAN be solved by hand with Cardano's method.
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
$endgroup$
The solution DO involves solving a cubic equation, which, by the way, CAN be solved by hand with Cardano's method.
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
answered Dec 6 '18 at 5:42
jjagmathjjagmath
2147
2147
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
add a comment |
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
Alternatively you could approximate it with 1 or 2 newton raphson iterations. With a suitable starting point like y = -3. I think the equation we are trying to solve is $(y + 5) ^ 3 + 8y = 0$
$endgroup$
– DanielOnMSE
Dec 6 '18 at 5:45
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
$begingroup$
I didn't know about cardano's method earlier. Thanks
$endgroup$
– Meet Shah
Dec 6 '18 at 5:58
add a comment |
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