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Find a point on an ellipse closest to a fixed point inside the ellipse


How to find a point on an ellipse whose normal intersects a point outside the ellipse?Solving lagrange multipliesSolving optimisation problem with langrangian methodUsing Lagrangian multipliers to check the solutionLagrangian multipliers in complex optimizationSolve $max_x_1,x_2,x_3 alpha min a x_1,b x_2,c x_3$ s.t. $p_1 x_1 + p_2 x_2 + p_3 x_3 = w$Does transformation of a function changes oprimal values?How to find Lagrange Multipliers in Quadratic Programming problem?Lagrangian multiplier of a Frobenius norm constraint optimization problemElement-wise Optimization with Lagrange Multiplier Vectorfinding extreme points for Lagrangian with multiple inequality constraints













2












$begingroup$


I want to find out $(u,v)$ on the ellipse $$fracu^2a^2+fracv^2b^2=1$$ for a point $(x,y)$ inside ellipse, which will denote shortest distance from $(x,y)$ to the ellipse boundary. I expressed this problem as a constrained optimization problem by this Lagrangian:



beginalign
L(u,v,lambda) &= (u-x)^2+(v-y)^2-lambdaleft(fracu^2a^2 + fracv^2b^2-1right)\[1ex]
fracpartial L(u,v,lambda)partial u&=2u-2x-frac2lambda ua^2\[1ex]
fracpartial L(u,v,lambda)partial v&=2v-2y-frac2lambda vb^2\[1ex]
fracpartial L(u,v,lambda)partial lambda&=fracu^2a^2+fracv^2b^2-1\[1em]
u&=fraca^2xa^2-lambdatag1\[1ex]
v&=fracb^2yb^2-lambdatag2\[1ex]
fracu^2a^2&+fracv^2b^2-1=0tag3\
endalign



After substituting $u$ and $v$ in equation $(3)$, I get
beginaligned
fraca^2x^2(a^2-lambda)^2+fracb^2y^2(b^2-lambda)^2-1=0\
endaligned

I am getting a 4th degree equation of $lambda$. I am facing difficulties in finding algebraic solution of the equation.










share|cite|improve this question











$endgroup$











  • $begingroup$
    I am afraid that any one would be facing the same difficulties.
    $endgroup$
    – Claude Leibovici
    Mar 30 at 4:17










  • $begingroup$
    Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
    $endgroup$
    – Ng Chung Tak
    Apr 1 at 10:38
















2












$begingroup$


I want to find out $(u,v)$ on the ellipse $$fracu^2a^2+fracv^2b^2=1$$ for a point $(x,y)$ inside ellipse, which will denote shortest distance from $(x,y)$ to the ellipse boundary. I expressed this problem as a constrained optimization problem by this Lagrangian:



beginalign
L(u,v,lambda) &= (u-x)^2+(v-y)^2-lambdaleft(fracu^2a^2 + fracv^2b^2-1right)\[1ex]
fracpartial L(u,v,lambda)partial u&=2u-2x-frac2lambda ua^2\[1ex]
fracpartial L(u,v,lambda)partial v&=2v-2y-frac2lambda vb^2\[1ex]
fracpartial L(u,v,lambda)partial lambda&=fracu^2a^2+fracv^2b^2-1\[1em]
u&=fraca^2xa^2-lambdatag1\[1ex]
v&=fracb^2yb^2-lambdatag2\[1ex]
fracu^2a^2&+fracv^2b^2-1=0tag3\
endalign



After substituting $u$ and $v$ in equation $(3)$, I get
beginaligned
fraca^2x^2(a^2-lambda)^2+fracb^2y^2(b^2-lambda)^2-1=0\
endaligned

I am getting a 4th degree equation of $lambda$. I am facing difficulties in finding algebraic solution of the equation.










share|cite|improve this question











$endgroup$











  • $begingroup$
    I am afraid that any one would be facing the same difficulties.
    $endgroup$
    – Claude Leibovici
    Mar 30 at 4:17










  • $begingroup$
    Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
    $endgroup$
    – Ng Chung Tak
    Apr 1 at 10:38














2












2








2





$begingroup$


I want to find out $(u,v)$ on the ellipse $$fracu^2a^2+fracv^2b^2=1$$ for a point $(x,y)$ inside ellipse, which will denote shortest distance from $(x,y)$ to the ellipse boundary. I expressed this problem as a constrained optimization problem by this Lagrangian:



beginalign
L(u,v,lambda) &= (u-x)^2+(v-y)^2-lambdaleft(fracu^2a^2 + fracv^2b^2-1right)\[1ex]
fracpartial L(u,v,lambda)partial u&=2u-2x-frac2lambda ua^2\[1ex]
fracpartial L(u,v,lambda)partial v&=2v-2y-frac2lambda vb^2\[1ex]
fracpartial L(u,v,lambda)partial lambda&=fracu^2a^2+fracv^2b^2-1\[1em]
u&=fraca^2xa^2-lambdatag1\[1ex]
v&=fracb^2yb^2-lambdatag2\[1ex]
fracu^2a^2&+fracv^2b^2-1=0tag3\
endalign



After substituting $u$ and $v$ in equation $(3)$, I get
beginaligned
fraca^2x^2(a^2-lambda)^2+fracb^2y^2(b^2-lambda)^2-1=0\
endaligned

I am getting a 4th degree equation of $lambda$. I am facing difficulties in finding algebraic solution of the equation.










share|cite|improve this question











$endgroup$




I want to find out $(u,v)$ on the ellipse $$fracu^2a^2+fracv^2b^2=1$$ for a point $(x,y)$ inside ellipse, which will denote shortest distance from $(x,y)$ to the ellipse boundary. I expressed this problem as a constrained optimization problem by this Lagrangian:



beginalign
L(u,v,lambda) &= (u-x)^2+(v-y)^2-lambdaleft(fracu^2a^2 + fracv^2b^2-1right)\[1ex]
fracpartial L(u,v,lambda)partial u&=2u-2x-frac2lambda ua^2\[1ex]
fracpartial L(u,v,lambda)partial v&=2v-2y-frac2lambda vb^2\[1ex]
fracpartial L(u,v,lambda)partial lambda&=fracu^2a^2+fracv^2b^2-1\[1em]
u&=fraca^2xa^2-lambdatag1\[1ex]
v&=fracb^2yb^2-lambdatag2\[1ex]
fracu^2a^2&+fracv^2b^2-1=0tag3\
endalign



After substituting $u$ and $v$ in equation $(3)$, I get
beginaligned
fraca^2x^2(a^2-lambda)^2+fracb^2y^2(b^2-lambda)^2-1=0\
endaligned

I am getting a 4th degree equation of $lambda$. I am facing difficulties in finding algebraic solution of the equation.







optimization lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 30 at 3:56









David M.

2,198421




2,198421










asked Mar 29 at 11:26









cseju19cseju19

215




215











  • $begingroup$
    I am afraid that any one would be facing the same difficulties.
    $endgroup$
    – Claude Leibovici
    Mar 30 at 4:17










  • $begingroup$
    Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
    $endgroup$
    – Ng Chung Tak
    Apr 1 at 10:38

















  • $begingroup$
    I am afraid that any one would be facing the same difficulties.
    $endgroup$
    – Claude Leibovici
    Mar 30 at 4:17










  • $begingroup$
    Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
    $endgroup$
    – Ng Chung Tak
    Apr 1 at 10:38
















$begingroup$
I am afraid that any one would be facing the same difficulties.
$endgroup$
– Claude Leibovici
Mar 30 at 4:17




$begingroup$
I am afraid that any one would be facing the same difficulties.
$endgroup$
– Claude Leibovici
Mar 30 at 4:17












$begingroup$
Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
$endgroup$
– Ng Chung Tak
Apr 1 at 10:38





$begingroup$
Your problem is equivalent to find a normal passing through a given point. There're two to four normals which concurrent to the given point. See another answer of mine here.
$endgroup$
– Ng Chung Tak
Apr 1 at 10:38











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