What to do further in Find the maximum area of ellipse…












1












$begingroup$


$$Question$$



Find the maximum area of ellipse that can be inscribed in an isosceles triangel of area $A$ and having one axis along the perpendiculur from the vertex of the triangle to the base.



$$Attempt$$



So isosceles triangle I made was of coordinates $D(frac{a} {2},p),B(0,0)$ and $C(a, 0)$.



Now Area of triangle =$A$=$frac{ap} {2}$ and the point E($frac{a} {2},0)$ as the feet of perpendicular on BC side by the vertex D.
So I assumed that this will be the major axis of that ellipse because the area of the ellipse is $pitimes atimes b$.



But, from here I am not able to do anything further.



Any suggestions or hints?



Thanks!






analytic-geometry triangle conic-sections area







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  • $begingroup$
    @symchdmath Please see now.
    $endgroup$
    – jayant98
    Dec 19 '18 at 10:16
















1












$begingroup$


$$Question$$



Find the maximum area of ellipse that can be inscribed in an isosceles triangel of area $A$ and having one axis along the perpendiculur from the vertex of the triangle to the base.



$$Attempt$$



So isosceles triangle I made was of coordinates $D(frac{a} {2},p),B(0,0)$ and $C(a, 0)$.



Now Area of triangle =$A$=$frac{ap} {2}$ and the point E($frac{a} {2},0)$ as the feet of perpendicular on BC side by the vertex D.
So I assumed that this will be the major axis of that ellipse because the area of the ellipse is $pitimes atimes b$.



But, from here I am not able to do anything further.



Any suggestions or hints?



Thanks!






analytic-geometry triangle conic-sections area







share|cite|improve this question











$endgroup$












  • $begingroup$
    @symchdmath Please see now.
    $endgroup$
    – jayant98
    Dec 19 '18 at 10:16














1












1








1





$begingroup$


$$Question$$



Find the maximum area of ellipse that can be inscribed in an isosceles triangel of area $A$ and having one axis along the perpendiculur from the vertex of the triangle to the base.



$$Attempt$$



So isosceles triangle I made was of coordinates $D(frac{a} {2},p),B(0,0)$ and $C(a, 0)$.



Now Area of triangle =$A$=$frac{ap} {2}$ and the point E($frac{a} {2},0)$ as the feet of perpendicular on BC side by the vertex D.
So I assumed that this will be the major axis of that ellipse because the area of the ellipse is $pitimes atimes b$.



But, from here I am not able to do anything further.



Any suggestions or hints?



Thanks!






analytic-geometry triangle conic-sections area







share|cite|improve this question











$endgroup$




$$Question$$



Find the maximum area of ellipse that can be inscribed in an isosceles triangel of area $A$ and having one axis along the perpendiculur from the vertex of the triangle to the base.



$$Attempt$$



So isosceles triangle I made was of coordinates $D(frac{a} {2},p),B(0,0)$ and $C(a, 0)$.



Now Area of triangle =$A$=$frac{ap} {2}$ and the point E($frac{a} {2},0)$ as the feet of perpendicular on BC side by the vertex D.
So I assumed that this will be the major axis of that ellipse because the area of the ellipse is $pitimes atimes b$.



But, from here I am not able to do anything further.



Any suggestions or hints?



Thanks!






analytic-geometry triangle conic-sections area




analytic-geometry triangle conic-sections area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 14:52







jayant98

















asked Dec 19 '18 at 9:56









jayant98jayant98

655318




655318












  • $begingroup$
    @symchdmath Please see now.
    $endgroup$
    – jayant98
    Dec 19 '18 at 10:16


















  • $begingroup$
    @symchdmath Please see now.
    $endgroup$
    – jayant98
    Dec 19 '18 at 10:16
















$begingroup$
@symchdmath Please see now.
$endgroup$
– jayant98
Dec 19 '18 at 10:16




$begingroup$
@symchdmath Please see now.
$endgroup$
– jayant98
Dec 19 '18 at 10:16










1 Answer
1






active

oldest

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7












$begingroup$

It's hard to solve this problem straightforward. However, there is a trick for you.



Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).



$$
R=frac{A_{incircle}}{A_{triangle}} = frac{pi r^2}{A_{triangle}}=frac{pi(2A_{triangle}/P)^2}{A_{triangle}}=4pifrac{A_{triangle}}{P^2}.
$$



We used here a formula $A_{triangle}=frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.



So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
    $endgroup$
    – jayant98
    Dec 19 '18 at 14:20













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

It's hard to solve this problem straightforward. However, there is a trick for you.



Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).



$$
R=frac{A_{incircle}}{A_{triangle}} = frac{pi r^2}{A_{triangle}}=frac{pi(2A_{triangle}/P)^2}{A_{triangle}}=4pifrac{A_{triangle}}{P^2}.
$$



We used here a formula $A_{triangle}=frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.



So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
    $endgroup$
    – jayant98
    Dec 19 '18 at 14:20


















7












$begingroup$

It's hard to solve this problem straightforward. However, there is a trick for you.



Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).



$$
R=frac{A_{incircle}}{A_{triangle}} = frac{pi r^2}{A_{triangle}}=frac{pi(2A_{triangle}/P)^2}{A_{triangle}}=4pifrac{A_{triangle}}{P^2}.
$$



We used here a formula $A_{triangle}=frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.



So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
    $endgroup$
    – jayant98
    Dec 19 '18 at 14:20
















7












7








7





$begingroup$

It's hard to solve this problem straightforward. However, there is a trick for you.



Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).



$$
R=frac{A_{incircle}}{A_{triangle}} = frac{pi r^2}{A_{triangle}}=frac{pi(2A_{triangle}/P)^2}{A_{triangle}}=4pifrac{A_{triangle}}{P^2}.
$$



We used here a formula $A_{triangle}=frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.



So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.






share|cite|improve this answer









$endgroup$



It's hard to solve this problem straightforward. However, there is a trick for you.



Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).



$$
R=frac{A_{incircle}}{A_{triangle}} = frac{pi r^2}{A_{triangle}}=frac{pi(2A_{triangle}/P)^2}{A_{triangle}}=4pifrac{A_{triangle}}{P^2}.
$$



We used here a formula $A_{triangle}=frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.



So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 12:11









Vasily MitchVasily Mitch

2,3141311




2,3141311








  • 1




    $begingroup$
    Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
    $endgroup$
    – jayant98
    Dec 19 '18 at 14:20
















  • 1




    $begingroup$
    Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
    $endgroup$
    – jayant98
    Dec 19 '18 at 14:20










1




1




$begingroup$
Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
$endgroup$
– jayant98
Dec 19 '18 at 14:20






$begingroup$
Or, is that the area of ellipse =$frac{(pi)A} {3sqrt{3}}$
$endgroup$
– jayant98
Dec 19 '18 at 14:20




















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