How to approach on this - finding minimum distance of point on the ellipse from the centre of it.
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Question
The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.
Attempt
Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.
Then how it is ellipse?
Any hints or suggestion?
analytic-geometry self-learning conic-sections
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
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add a comment |
$begingroup$
Question
The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.
Attempt
Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.
Then how it is ellipse?
Any hints or suggestion?
analytic-geometry self-learning conic-sections
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02
add a comment |
$begingroup$
Question
The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.
Attempt
Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.
Then how it is ellipse?
Any hints or suggestion?
analytic-geometry self-learning conic-sections
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
$endgroup$
Question
The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.
Attempt
Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.
Then how it is ellipse?
Any hints or suggestion?
analytic-geometry self-learning conic-sections
analytic-geometry self-learning conic-sections
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
edited Dec 18 '18 at 9:07
jayant98
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
asked Dec 17 '18 at 18:09
jayant98jayant98
653318
653318
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02
add a comment |
2 Answers
2
active
oldest
votes
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The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
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add a comment |
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The matrix
$$
begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
$$
defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
$$
left{begin{array}{l}
x^2+4xy+3y^2=4 \
(x,y) = t(sqrt5-1,2) \
end{array}right.
$$
The solutions are
$$
x_0 = sqrt{frac{14}{sqrt{5
}}-6}
qquad
y_0 = sqrt{frac{6}{sqrt{
5}}-2}
$$
and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is
$$
|(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
$$
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
$endgroup$
The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic
Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
edited Dec 17 '18 at 19:48
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
answered Dec 17 '18 at 19:11
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
$begingroup$
The matrix
$$
begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
$$
defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
$$
left{begin{array}{l}
x^2+4xy+3y^2=4 \
(x,y) = t(sqrt5-1,2) \
end{array}right.
$$
The solutions are
$$
x_0 = sqrt{frac{14}{sqrt{5
}}-6}
qquad
y_0 = sqrt{frac{6}{sqrt{
5}}-2}
$$
and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is
$$
|(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
$$
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
$endgroup$
add a comment |
$begingroup$
The matrix
$$
begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
$$
defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
$$
left{begin{array}{l}
x^2+4xy+3y^2=4 \
(x,y) = t(sqrt5-1,2) \
end{array}right.
$$
The solutions are
$$
x_0 = sqrt{frac{14}{sqrt{5
}}-6}
qquad
y_0 = sqrt{frac{6}{sqrt{
5}}-2}
$$
and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is
$$
|(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
$$
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
$endgroup$
add a comment |
$begingroup$
The matrix
$$
begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
$$
defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
$$
left{begin{array}{l}
x^2+4xy+3y^2=4 \
(x,y) = t(sqrt5-1,2) \
end{array}right.
$$
The solutions are
$$
x_0 = sqrt{frac{14}{sqrt{5
}}-6}
qquad
y_0 = sqrt{frac{6}{sqrt{
5}}-2}
$$
and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is
$$
|(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
$$
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
$endgroup$
The matrix
$$
begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
$$
defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
$$
left{begin{array}{l}
x^2+4xy+3y^2=4 \
(x,y) = t(sqrt5-1,2) \
end{array}right.
$$
The solutions are
$$
x_0 = sqrt{frac{14}{sqrt{5
}}-6}
qquad
y_0 = sqrt{frac{6}{sqrt{
5}}-2}
$$
and the opposite point $-(x_0,y_0)$.
Therefore the distance from the center (which is the origin) is
$$
|(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
$$
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
answered Dec 17 '18 at 19:25
FedericoFederico
5,144514
5,144514
add a comment |
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Dec 19 '18 at 2:02